Let $sum_{k=1}^infty a_n$ be convergent show that $sum_{k=1}^infty n(a_n-a_{n+1})$ converges












5












$begingroup$


Let $sumlimits_{k=1}^infty a_n$ be a convergent series where $a_ngeq0$ and $(a_n)$ is a monotone decreasing sequence prove that the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges.



What I tried :



Let $(A_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty a_n$ and $(B_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$.



Since $A_n=sumlimits_{k=1}^n a_k$ and $B_n=sumlimits_{k=1}^n k(a_k-a_{k+1})$ we get that:



begin{align}
B_n&=A_n-na_{n+1}\
&=(a_1-a_{n+1})+(a_2-a_{n+1})+...+(a_n-a_{n+1})\
&>(a_1-a_n)+(a_2-a_n)+...+(a_n-a_n)\
&=B_{n-1}
end{align}



we see that $(B_n)$ is a monotone increasing sequence $...(1)$



and



$B_n=A_n-na_{n+1}<A_n$ this implies that the sequence $(B_n)$ is bounded above ...(2)



Therefore (from (1) and (2)) the sequence $(B_n)$ converges so the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges



Is my proof correct?






real-analysis sequences-and-series proof-verification convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Looks ok. Some simplifications: you know that $B_n - B_{n-1} = n(a_n - a_{n+1})$ which is $>0$. Another way to look at it: you know that $A_n$ converges and $B_n = A_n - na_{n+1}$. This means that the converge of $B_n$ is equivalent to showing $na_{n+1} to 0$ as $ntoinfty$.
    $endgroup$
    – Winther
    Dec 25 '18 at 12:48












  • $begingroup$
    Thank you , and from the second one do we get that the sum of $sum_{k=1}^infty n(a_n-a_{n+1})$ equals the sum of $sum_{k=1}^infty a_n$?
    $endgroup$
    – Maths Survivor
    Dec 25 '18 at 13:00








  • 1




    $begingroup$
    Taking the limit $ntoinfty$ of $B_n = A_n - na_{n+1}$. However disregard the last thing I said. The usual way of showing that $na_{n+1} to 0$ is to show that $sum n(a_n-a_{n+1})$ converges so that would be circular (see e.g. math.stackexchange.com/questions/383769/…). Your approach is good. btw that $B_n = A_n - na_{n+1}$ is a special case of summation by parts.
    $endgroup$
    – Winther
    Dec 25 '18 at 13:02












  • $begingroup$
    I can use another method to prove that $na_nto0$ when $n to infty$ without using the convergence of the series $sum_{k=1}^infty n(a_n-a_{n+1})$ , so from $B_n=A_n-na_n$ and since the sequences $(A_n)$ and $(na_n)$ converge when $n to infty $ implies that also the sequnce $(B_n)$ converges right?
    $endgroup$
    – Maths Survivor
    Dec 25 '18 at 13:18










  • $begingroup$
    Yes that's right
    $endgroup$
    – Winther
    Dec 25 '18 at 13:19
















5












$begingroup$


Let $sumlimits_{k=1}^infty a_n$ be a convergent series where $a_ngeq0$ and $(a_n)$ is a monotone decreasing sequence prove that the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges.



What I tried :



Let $(A_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty a_n$ and $(B_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$.



Since $A_n=sumlimits_{k=1}^n a_k$ and $B_n=sumlimits_{k=1}^n k(a_k-a_{k+1})$ we get that:



begin{align}
B_n&=A_n-na_{n+1}\
&=(a_1-a_{n+1})+(a_2-a_{n+1})+...+(a_n-a_{n+1})\
&>(a_1-a_n)+(a_2-a_n)+...+(a_n-a_n)\
&=B_{n-1}
end{align}



we see that $(B_n)$ is a monotone increasing sequence $...(1)$



and



$B_n=A_n-na_{n+1}<A_n$ this implies that the sequence $(B_n)$ is bounded above ...(2)



Therefore (from (1) and (2)) the sequence $(B_n)$ converges so the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges



Is my proof correct?






real-analysis sequences-and-series proof-verification convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Looks ok. Some simplifications: you know that $B_n - B_{n-1} = n(a_n - a_{n+1})$ which is $>0$. Another way to look at it: you know that $A_n$ converges and $B_n = A_n - na_{n+1}$. This means that the converge of $B_n$ is equivalent to showing $na_{n+1} to 0$ as $ntoinfty$.
    $endgroup$
    – Winther
    Dec 25 '18 at 12:48












  • $begingroup$
    Thank you , and from the second one do we get that the sum of $sum_{k=1}^infty n(a_n-a_{n+1})$ equals the sum of $sum_{k=1}^infty a_n$?
    $endgroup$
    – Maths Survivor
    Dec 25 '18 at 13:00








  • 1




    $begingroup$
    Taking the limit $ntoinfty$ of $B_n = A_n - na_{n+1}$. However disregard the last thing I said. The usual way of showing that $na_{n+1} to 0$ is to show that $sum n(a_n-a_{n+1})$ converges so that would be circular (see e.g. math.stackexchange.com/questions/383769/…). Your approach is good. btw that $B_n = A_n - na_{n+1}$ is a special case of summation by parts.
    $endgroup$
    – Winther
    Dec 25 '18 at 13:02












  • $begingroup$
    I can use another method to prove that $na_nto0$ when $n to infty$ without using the convergence of the series $sum_{k=1}^infty n(a_n-a_{n+1})$ , so from $B_n=A_n-na_n$ and since the sequences $(A_n)$ and $(na_n)$ converge when $n to infty $ implies that also the sequnce $(B_n)$ converges right?
    $endgroup$
    – Maths Survivor
    Dec 25 '18 at 13:18










  • $begingroup$
    Yes that's right
    $endgroup$
    – Winther
    Dec 25 '18 at 13:19














5












5








5


2



$begingroup$


Let $sumlimits_{k=1}^infty a_n$ be a convergent series where $a_ngeq0$ and $(a_n)$ is a monotone decreasing sequence prove that the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges.



What I tried :



Let $(A_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty a_n$ and $(B_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$.



Since $A_n=sumlimits_{k=1}^n a_k$ and $B_n=sumlimits_{k=1}^n k(a_k-a_{k+1})$ we get that:



begin{align}
B_n&=A_n-na_{n+1}\
&=(a_1-a_{n+1})+(a_2-a_{n+1})+...+(a_n-a_{n+1})\
&>(a_1-a_n)+(a_2-a_n)+...+(a_n-a_n)\
&=B_{n-1}
end{align}



we see that $(B_n)$ is a monotone increasing sequence $...(1)$



and



$B_n=A_n-na_{n+1}<A_n$ this implies that the sequence $(B_n)$ is bounded above ...(2)



Therefore (from (1) and (2)) the sequence $(B_n)$ converges so the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges



Is my proof correct?






real-analysis sequences-and-series proof-verification convergence







share|cite|improve this question











$endgroup$




Let $sumlimits_{k=1}^infty a_n$ be a convergent series where $a_ngeq0$ and $(a_n)$ is a monotone decreasing sequence prove that the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges.



What I tried :



Let $(A_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty a_n$ and $(B_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$.



Since $A_n=sumlimits_{k=1}^n a_k$ and $B_n=sumlimits_{k=1}^n k(a_k-a_{k+1})$ we get that:



begin{align}
B_n&=A_n-na_{n+1}\
&=(a_1-a_{n+1})+(a_2-a_{n+1})+...+(a_n-a_{n+1})\
&>(a_1-a_n)+(a_2-a_n)+...+(a_n-a_n)\
&=B_{n-1}
end{align}



we see that $(B_n)$ is a monotone increasing sequence $...(1)$



and



$B_n=A_n-na_{n+1}<A_n$ this implies that the sequence $(B_n)$ is bounded above ...(2)



Therefore (from (1) and (2)) the sequence $(B_n)$ converges so the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges



Is my proof correct?






real-analysis sequences-and-series proof-verification convergence




real-analysis sequences-and-series proof-verification convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 17:12









Henning Makholm

241k17308549




241k17308549










asked Dec 25 '18 at 12:34









Maths SurvivorMaths Survivor

502219




502219












  • $begingroup$
    Looks ok. Some simplifications: you know that $B_n - B_{n-1} = n(a_n - a_{n+1})$ which is $>0$. Another way to look at it: you know that $A_n$ converges and $B_n = A_n - na_{n+1}$. This means that the converge of $B_n$ is equivalent to showing $na_{n+1} to 0$ as $ntoinfty$.
    $endgroup$
    – Winther
    Dec 25 '18 at 12:48












  • $begingroup$
    Thank you , and from the second one do we get that the sum of $sum_{k=1}^infty n(a_n-a_{n+1})$ equals the sum of $sum_{k=1}^infty a_n$?
    $endgroup$
    – Maths Survivor
    Dec 25 '18 at 13:00








  • 1




    $begingroup$
    Taking the limit $ntoinfty$ of $B_n = A_n - na_{n+1}$. However disregard the last thing I said. The usual way of showing that $na_{n+1} to 0$ is to show that $sum n(a_n-a_{n+1})$ converges so that would be circular (see e.g. math.stackexchange.com/questions/383769/…). Your approach is good. btw that $B_n = A_n - na_{n+1}$ is a special case of summation by parts.
    $endgroup$
    – Winther
    Dec 25 '18 at 13:02












  • $begingroup$
    I can use another method to prove that $na_nto0$ when $n to infty$ without using the convergence of the series $sum_{k=1}^infty n(a_n-a_{n+1})$ , so from $B_n=A_n-na_n$ and since the sequences $(A_n)$ and $(na_n)$ converge when $n to infty $ implies that also the sequnce $(B_n)$ converges right?
    $endgroup$
    – Maths Survivor
    Dec 25 '18 at 13:18










  • $begingroup$
    Yes that's right
    $endgroup$
    – Winther
    Dec 25 '18 at 13:19


















  • $begingroup$
    Looks ok. Some simplifications: you know that $B_n - B_{n-1} = n(a_n - a_{n+1})$ which is $>0$. Another way to look at it: you know that $A_n$ converges and $B_n = A_n - na_{n+1}$. This means that the converge of $B_n$ is equivalent to showing $na_{n+1} to 0$ as $ntoinfty$.
    $endgroup$
    – Winther
    Dec 25 '18 at 12:48












  • $begingroup$
    Thank you , and from the second one do we get that the sum of $sum_{k=1}^infty n(a_n-a_{n+1})$ equals the sum of $sum_{k=1}^infty a_n$?
    $endgroup$
    – Maths Survivor
    Dec 25 '18 at 13:00








  • 1




    $begingroup$
    Taking the limit $ntoinfty$ of $B_n = A_n - na_{n+1}$. However disregard the last thing I said. The usual way of showing that $na_{n+1} to 0$ is to show that $sum n(a_n-a_{n+1})$ converges so that would be circular (see e.g. math.stackexchange.com/questions/383769/…). Your approach is good. btw that $B_n = A_n - na_{n+1}$ is a special case of summation by parts.
    $endgroup$
    – Winther
    Dec 25 '18 at 13:02












  • $begingroup$
    I can use another method to prove that $na_nto0$ when $n to infty$ without using the convergence of the series $sum_{k=1}^infty n(a_n-a_{n+1})$ , so from $B_n=A_n-na_n$ and since the sequences $(A_n)$ and $(na_n)$ converge when $n to infty $ implies that also the sequnce $(B_n)$ converges right?
    $endgroup$
    – Maths Survivor
    Dec 25 '18 at 13:18










  • $begingroup$
    Yes that's right
    $endgroup$
    – Winther
    Dec 25 '18 at 13:19
















$begingroup$
Looks ok. Some simplifications: you know that $B_n - B_{n-1} = n(a_n - a_{n+1})$ which is $>0$. Another way to look at it: you know that $A_n$ converges and $B_n = A_n - na_{n+1}$. This means that the converge of $B_n$ is equivalent to showing $na_{n+1} to 0$ as $ntoinfty$.
$endgroup$
– Winther
Dec 25 '18 at 12:48






$begingroup$
Looks ok. Some simplifications: you know that $B_n - B_{n-1} = n(a_n - a_{n+1})$ which is $>0$. Another way to look at it: you know that $A_n$ converges and $B_n = A_n - na_{n+1}$. This means that the converge of $B_n$ is equivalent to showing $na_{n+1} to 0$ as $ntoinfty$.
$endgroup$
– Winther
Dec 25 '18 at 12:48














$begingroup$
Thank you , and from the second one do we get that the sum of $sum_{k=1}^infty n(a_n-a_{n+1})$ equals the sum of $sum_{k=1}^infty a_n$?
$endgroup$
– Maths Survivor
Dec 25 '18 at 13:00






$begingroup$
Thank you , and from the second one do we get that the sum of $sum_{k=1}^infty n(a_n-a_{n+1})$ equals the sum of $sum_{k=1}^infty a_n$?
$endgroup$
– Maths Survivor
Dec 25 '18 at 13:00






1




1




$begingroup$
Taking the limit $ntoinfty$ of $B_n = A_n - na_{n+1}$. However disregard the last thing I said. The usual way of showing that $na_{n+1} to 0$ is to show that $sum n(a_n-a_{n+1})$ converges so that would be circular (see e.g. math.stackexchange.com/questions/383769/…). Your approach is good. btw that $B_n = A_n - na_{n+1}$ is a special case of summation by parts.
$endgroup$
– Winther
Dec 25 '18 at 13:02






$begingroup$
Taking the limit $ntoinfty$ of $B_n = A_n - na_{n+1}$. However disregard the last thing I said. The usual way of showing that $na_{n+1} to 0$ is to show that $sum n(a_n-a_{n+1})$ converges so that would be circular (see e.g. math.stackexchange.com/questions/383769/…). Your approach is good. btw that $B_n = A_n - na_{n+1}$ is a special case of summation by parts.
$endgroup$
– Winther
Dec 25 '18 at 13:02














$begingroup$
I can use another method to prove that $na_nto0$ when $n to infty$ without using the convergence of the series $sum_{k=1}^infty n(a_n-a_{n+1})$ , so from $B_n=A_n-na_n$ and since the sequences $(A_n)$ and $(na_n)$ converge when $n to infty $ implies that also the sequnce $(B_n)$ converges right?
$endgroup$
– Maths Survivor
Dec 25 '18 at 13:18




$begingroup$
I can use another method to prove that $na_nto0$ when $n to infty$ without using the convergence of the series $sum_{k=1}^infty n(a_n-a_{n+1})$ , so from $B_n=A_n-na_n$ and since the sequences $(A_n)$ and $(na_n)$ converge when $n to infty $ implies that also the sequnce $(B_n)$ converges right?
$endgroup$
– Maths Survivor
Dec 25 '18 at 13:18












$begingroup$
Yes that's right
$endgroup$
– Winther
Dec 25 '18 at 13:19




$begingroup$
Yes that's right
$endgroup$
– Winther
Dec 25 '18 at 13:19










2 Answers
2






active

oldest

votes


















2












$begingroup$

Some comments:




  1. The identity $B_n = A_n - n a_{n+1}$ is not quite obvious enough to state without proof. A quick induction proof would work, as would writing out the sums using ellipses (i.e. the symbol "$ldots$") and simplifying.


  2. Showing $B_n < A_n$ establishes an upper bound based on $n$, which is not allowed (e.g. $B_n le B_n$ always!). As $A_n$ (being the partial sums of a convergent series) is convergent, you can easily establish an upper bound (especially when you consider the fact that $A_n$ is increasing).


  3. You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.


  4. You could also further note that, if $lim B_n < lim A_n$, then $a_n$ is approximately a multiple of the harmonic series, which is divergent, thus the two series share the same sum. (This is not a criticism, just something worth noting).







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    "then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence?
    $endgroup$
    – Maths Survivor
    Dec 25 '18 at 13:12






  • 1




    $begingroup$
    @MathsSurvivor If $A_n to A$ and $B_n to B$ with $A neq B$, then $$frac{a_{n+1}}{frac{1}{n}} = na_{n+1} = A_n - B_n to A - B neq 0,$$ so by the limit comparison test, $a_{n+1}$, when summed, is a divergent sequence. This contradicts $A_n$ converging.
    $endgroup$
    – Theo Bendit
    Dec 25 '18 at 14:02





















2












$begingroup$

Note that we can use $n=sum_{k=1}^n (1)$ to write



$$begin{align}
sum_{n=1}^N n(a_{n+1}-a_n)&=sum_{n=1}^N sum_{k=1}^n(a_{n+1}-a_n)\\
&=sum_{k=1}^N sum_{n=k}^N (a_{n+1}-a_n)\\
&=sum_{k=1}^N (a_{N+1}-a_k)\\
&=Na_{N+1}-sum_{k=1}^N a_ktag1
end{align}$$



Inasmuch as $a_nge 0$ monotonically decreases to $0$, and $sum_{k=1}^n a_n<infty$, we have $lim_{ntoinfty }na_n=0$. Hence, using $(1)$, we see that



$$begin{align}
lim_{Nto infty }sum_{n=1}^N n(a_{n+1}-a_n)&=lim_{Ntoinfty}left(Na_{N+1}-sum_{k=1}^N a_kright)\\
&=-sum_{n=1}^infty a_n
end{align}$$



from which we conclude that $sum_{n=1}^infty n(a_{n+1}-a_n)$ converges.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Winther's comment on the question suggests that showing $lim n a_n = 0$ is often done by first establishing $sum n(a_n - a_{n+1}) = sum a_n$. Out of curiosity, how would you establish $lim n a_n = 0$?
    $endgroup$
    – Theo Bendit
    Dec 25 '18 at 22:44












  • $begingroup$
    Note that $$(2n)a_{2n}le 2sum_{n+1}^{2n}a_kto 0$$since $a_nge0$ is monotonic and $sum_n a_n<infty.$.
    $endgroup$
    – Mark Viola
    Dec 26 '18 at 4:05













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Some comments:




  1. The identity $B_n = A_n - n a_{n+1}$ is not quite obvious enough to state without proof. A quick induction proof would work, as would writing out the sums using ellipses (i.e. the symbol "$ldots$") and simplifying.


  2. Showing $B_n < A_n$ establishes an upper bound based on $n$, which is not allowed (e.g. $B_n le B_n$ always!). As $A_n$ (being the partial sums of a convergent series) is convergent, you can easily establish an upper bound (especially when you consider the fact that $A_n$ is increasing).


  3. You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.


  4. You could also further note that, if $lim B_n < lim A_n$, then $a_n$ is approximately a multiple of the harmonic series, which is divergent, thus the two series share the same sum. (This is not a criticism, just something worth noting).







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    "then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence?
    $endgroup$
    – Maths Survivor
    Dec 25 '18 at 13:12






  • 1




    $begingroup$
    @MathsSurvivor If $A_n to A$ and $B_n to B$ with $A neq B$, then $$frac{a_{n+1}}{frac{1}{n}} = na_{n+1} = A_n - B_n to A - B neq 0,$$ so by the limit comparison test, $a_{n+1}$, when summed, is a divergent sequence. This contradicts $A_n$ converging.
    $endgroup$
    – Theo Bendit
    Dec 25 '18 at 14:02


















2












$begingroup$

Some comments:




  1. The identity $B_n = A_n - n a_{n+1}$ is not quite obvious enough to state without proof. A quick induction proof would work, as would writing out the sums using ellipses (i.e. the symbol "$ldots$") and simplifying.


  2. Showing $B_n < A_n$ establishes an upper bound based on $n$, which is not allowed (e.g. $B_n le B_n$ always!). As $A_n$ (being the partial sums of a convergent series) is convergent, you can easily establish an upper bound (especially when you consider the fact that $A_n$ is increasing).


  3. You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.


  4. You could also further note that, if $lim B_n < lim A_n$, then $a_n$ is approximately a multiple of the harmonic series, which is divergent, thus the two series share the same sum. (This is not a criticism, just something worth noting).







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    "then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence?
    $endgroup$
    – Maths Survivor
    Dec 25 '18 at 13:12






  • 1




    $begingroup$
    @MathsSurvivor If $A_n to A$ and $B_n to B$ with $A neq B$, then $$frac{a_{n+1}}{frac{1}{n}} = na_{n+1} = A_n - B_n to A - B neq 0,$$ so by the limit comparison test, $a_{n+1}$, when summed, is a divergent sequence. This contradicts $A_n$ converging.
    $endgroup$
    – Theo Bendit
    Dec 25 '18 at 14:02
















2












2








2





$begingroup$

Some comments:




  1. The identity $B_n = A_n - n a_{n+1}$ is not quite obvious enough to state without proof. A quick induction proof would work, as would writing out the sums using ellipses (i.e. the symbol "$ldots$") and simplifying.


  2. Showing $B_n < A_n$ establishes an upper bound based on $n$, which is not allowed (e.g. $B_n le B_n$ always!). As $A_n$ (being the partial sums of a convergent series) is convergent, you can easily establish an upper bound (especially when you consider the fact that $A_n$ is increasing).


  3. You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.


  4. You could also further note that, if $lim B_n < lim A_n$, then $a_n$ is approximately a multiple of the harmonic series, which is divergent, thus the two series share the same sum. (This is not a criticism, just something worth noting).







share|cite|improve this answer









$endgroup$



Some comments:




  1. The identity $B_n = A_n - n a_{n+1}$ is not quite obvious enough to state without proof. A quick induction proof would work, as would writing out the sums using ellipses (i.e. the symbol "$ldots$") and simplifying.


  2. Showing $B_n < A_n$ establishes an upper bound based on $n$, which is not allowed (e.g. $B_n le B_n$ always!). As $A_n$ (being the partial sums of a convergent series) is convergent, you can easily establish an upper bound (especially when you consider the fact that $A_n$ is increasing).


  3. You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.


  4. You could also further note that, if $lim B_n < lim A_n$, then $a_n$ is approximately a multiple of the harmonic series, which is divergent, thus the two series share the same sum. (This is not a criticism, just something worth noting).








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 25 '18 at 13:05









Theo BenditTheo Bendit

19.4k12353




19.4k12353








  • 1




    $begingroup$
    "then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence?
    $endgroup$
    – Maths Survivor
    Dec 25 '18 at 13:12






  • 1




    $begingroup$
    @MathsSurvivor If $A_n to A$ and $B_n to B$ with $A neq B$, then $$frac{a_{n+1}}{frac{1}{n}} = na_{n+1} = A_n - B_n to A - B neq 0,$$ so by the limit comparison test, $a_{n+1}$, when summed, is a divergent sequence. This contradicts $A_n$ converging.
    $endgroup$
    – Theo Bendit
    Dec 25 '18 at 14:02
















  • 1




    $begingroup$
    "then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence?
    $endgroup$
    – Maths Survivor
    Dec 25 '18 at 13:12






  • 1




    $begingroup$
    @MathsSurvivor If $A_n to A$ and $B_n to B$ with $A neq B$, then $$frac{a_{n+1}}{frac{1}{n}} = na_{n+1} = A_n - B_n to A - B neq 0,$$ so by the limit comparison test, $a_{n+1}$, when summed, is a divergent sequence. This contradicts $A_n$ converging.
    $endgroup$
    – Theo Bendit
    Dec 25 '18 at 14:02










1




1




$begingroup$
"then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence?
$endgroup$
– Maths Survivor
Dec 25 '18 at 13:12




$begingroup$
"then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence?
$endgroup$
– Maths Survivor
Dec 25 '18 at 13:12




1




1




$begingroup$
@MathsSurvivor If $A_n to A$ and $B_n to B$ with $A neq B$, then $$frac{a_{n+1}}{frac{1}{n}} = na_{n+1} = A_n - B_n to A - B neq 0,$$ so by the limit comparison test, $a_{n+1}$, when summed, is a divergent sequence. This contradicts $A_n$ converging.
$endgroup$
– Theo Bendit
Dec 25 '18 at 14:02






$begingroup$
@MathsSurvivor If $A_n to A$ and $B_n to B$ with $A neq B$, then $$frac{a_{n+1}}{frac{1}{n}} = na_{n+1} = A_n - B_n to A - B neq 0,$$ so by the limit comparison test, $a_{n+1}$, when summed, is a divergent sequence. This contradicts $A_n$ converging.
$endgroup$
– Theo Bendit
Dec 25 '18 at 14:02













2












$begingroup$

Note that we can use $n=sum_{k=1}^n (1)$ to write



$$begin{align}
sum_{n=1}^N n(a_{n+1}-a_n)&=sum_{n=1}^N sum_{k=1}^n(a_{n+1}-a_n)\\
&=sum_{k=1}^N sum_{n=k}^N (a_{n+1}-a_n)\\
&=sum_{k=1}^N (a_{N+1}-a_k)\\
&=Na_{N+1}-sum_{k=1}^N a_ktag1
end{align}$$



Inasmuch as $a_nge 0$ monotonically decreases to $0$, and $sum_{k=1}^n a_n<infty$, we have $lim_{ntoinfty }na_n=0$. Hence, using $(1)$, we see that



$$begin{align}
lim_{Nto infty }sum_{n=1}^N n(a_{n+1}-a_n)&=lim_{Ntoinfty}left(Na_{N+1}-sum_{k=1}^N a_kright)\\
&=-sum_{n=1}^infty a_n
end{align}$$



from which we conclude that $sum_{n=1}^infty n(a_{n+1}-a_n)$ converges.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Winther's comment on the question suggests that showing $lim n a_n = 0$ is often done by first establishing $sum n(a_n - a_{n+1}) = sum a_n$. Out of curiosity, how would you establish $lim n a_n = 0$?
    $endgroup$
    – Theo Bendit
    Dec 25 '18 at 22:44












  • $begingroup$
    Note that $$(2n)a_{2n}le 2sum_{n+1}^{2n}a_kto 0$$since $a_nge0$ is monotonic and $sum_n a_n<infty.$.
    $endgroup$
    – Mark Viola
    Dec 26 '18 at 4:05


















2












$begingroup$

Note that we can use $n=sum_{k=1}^n (1)$ to write



$$begin{align}
sum_{n=1}^N n(a_{n+1}-a_n)&=sum_{n=1}^N sum_{k=1}^n(a_{n+1}-a_n)\\
&=sum_{k=1}^N sum_{n=k}^N (a_{n+1}-a_n)\\
&=sum_{k=1}^N (a_{N+1}-a_k)\\
&=Na_{N+1}-sum_{k=1}^N a_ktag1
end{align}$$



Inasmuch as $a_nge 0$ monotonically decreases to $0$, and $sum_{k=1}^n a_n<infty$, we have $lim_{ntoinfty }na_n=0$. Hence, using $(1)$, we see that



$$begin{align}
lim_{Nto infty }sum_{n=1}^N n(a_{n+1}-a_n)&=lim_{Ntoinfty}left(Na_{N+1}-sum_{k=1}^N a_kright)\\
&=-sum_{n=1}^infty a_n
end{align}$$



from which we conclude that $sum_{n=1}^infty n(a_{n+1}-a_n)$ converges.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Winther's comment on the question suggests that showing $lim n a_n = 0$ is often done by first establishing $sum n(a_n - a_{n+1}) = sum a_n$. Out of curiosity, how would you establish $lim n a_n = 0$?
    $endgroup$
    – Theo Bendit
    Dec 25 '18 at 22:44












  • $begingroup$
    Note that $$(2n)a_{2n}le 2sum_{n+1}^{2n}a_kto 0$$since $a_nge0$ is monotonic and $sum_n a_n<infty.$.
    $endgroup$
    – Mark Viola
    Dec 26 '18 at 4:05
















2












2








2





$begingroup$

Note that we can use $n=sum_{k=1}^n (1)$ to write



$$begin{align}
sum_{n=1}^N n(a_{n+1}-a_n)&=sum_{n=1}^N sum_{k=1}^n(a_{n+1}-a_n)\\
&=sum_{k=1}^N sum_{n=k}^N (a_{n+1}-a_n)\\
&=sum_{k=1}^N (a_{N+1}-a_k)\\
&=Na_{N+1}-sum_{k=1}^N a_ktag1
end{align}$$



Inasmuch as $a_nge 0$ monotonically decreases to $0$, and $sum_{k=1}^n a_n<infty$, we have $lim_{ntoinfty }na_n=0$. Hence, using $(1)$, we see that



$$begin{align}
lim_{Nto infty }sum_{n=1}^N n(a_{n+1}-a_n)&=lim_{Ntoinfty}left(Na_{N+1}-sum_{k=1}^N a_kright)\\
&=-sum_{n=1}^infty a_n
end{align}$$



from which we conclude that $sum_{n=1}^infty n(a_{n+1}-a_n)$ converges.






share|cite|improve this answer









$endgroup$



Note that we can use $n=sum_{k=1}^n (1)$ to write



$$begin{align}
sum_{n=1}^N n(a_{n+1}-a_n)&=sum_{n=1}^N sum_{k=1}^n(a_{n+1}-a_n)\\
&=sum_{k=1}^N sum_{n=k}^N (a_{n+1}-a_n)\\
&=sum_{k=1}^N (a_{N+1}-a_k)\\
&=Na_{N+1}-sum_{k=1}^N a_ktag1
end{align}$$



Inasmuch as $a_nge 0$ monotonically decreases to $0$, and $sum_{k=1}^n a_n<infty$, we have $lim_{ntoinfty }na_n=0$. Hence, using $(1)$, we see that



$$begin{align}
lim_{Nto infty }sum_{n=1}^N n(a_{n+1}-a_n)&=lim_{Ntoinfty}left(Na_{N+1}-sum_{k=1}^N a_kright)\\
&=-sum_{n=1}^infty a_n
end{align}$$



from which we conclude that $sum_{n=1}^infty n(a_{n+1}-a_n)$ converges.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 25 '18 at 17:11









Mark ViolaMark Viola

133k1277176




133k1277176












  • $begingroup$
    Winther's comment on the question suggests that showing $lim n a_n = 0$ is often done by first establishing $sum n(a_n - a_{n+1}) = sum a_n$. Out of curiosity, how would you establish $lim n a_n = 0$?
    $endgroup$
    – Theo Bendit
    Dec 25 '18 at 22:44












  • $begingroup$
    Note that $$(2n)a_{2n}le 2sum_{n+1}^{2n}a_kto 0$$since $a_nge0$ is monotonic and $sum_n a_n<infty.$.
    $endgroup$
    – Mark Viola
    Dec 26 '18 at 4:05




















  • $begingroup$
    Winther's comment on the question suggests that showing $lim n a_n = 0$ is often done by first establishing $sum n(a_n - a_{n+1}) = sum a_n$. Out of curiosity, how would you establish $lim n a_n = 0$?
    $endgroup$
    – Theo Bendit
    Dec 25 '18 at 22:44












  • $begingroup$
    Note that $$(2n)a_{2n}le 2sum_{n+1}^{2n}a_kto 0$$since $a_nge0$ is monotonic and $sum_n a_n<infty.$.
    $endgroup$
    – Mark Viola
    Dec 26 '18 at 4:05


















$begingroup$
Winther's comment on the question suggests that showing $lim n a_n = 0$ is often done by first establishing $sum n(a_n - a_{n+1}) = sum a_n$. Out of curiosity, how would you establish $lim n a_n = 0$?
$endgroup$
– Theo Bendit
Dec 25 '18 at 22:44






$begingroup$
Winther's comment on the question suggests that showing $lim n a_n = 0$ is often done by first establishing $sum n(a_n - a_{n+1}) = sum a_n$. Out of curiosity, how would you establish $lim n a_n = 0$?
$endgroup$
– Theo Bendit
Dec 25 '18 at 22:44














$begingroup$
Note that $$(2n)a_{2n}le 2sum_{n+1}^{2n}a_kto 0$$since $a_nge0$ is monotonic and $sum_n a_n<infty.$.
$endgroup$
– Mark Viola
Dec 26 '18 at 4:05






$begingroup$
Note that $$(2n)a_{2n}le 2sum_{n+1}^{2n}a_kto 0$$since $a_nge0$ is monotonic and $sum_n a_n<infty.$.
$endgroup$
– Mark Viola
Dec 26 '18 at 4:05




















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