Let $sum_{k=1}^infty a_n$ be convergent show that $sum_{k=1}^infty n(a_n-a_{n+1})$ converges
$begingroup$
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
edited Dec 25 '18 at 17:12
Henning Makholm
241k17308549
asked Dec 25 '18 at 12:34
Maths SurvivorMaths Survivor
502219
$endgroup$
|
show 3 more comments
$begingroup$
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
edited Dec 25 '18 at 17:12
Henning Makholm
241k17308549
asked Dec 25 '18 at 12:34
Maths SurvivorMaths Survivor
502219
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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
edited Dec 25 '18 at 17:12
Henning Makholm
241k17308549
asked Dec 25 '18 at 12:34
Maths SurvivorMaths Survivor
502219
$endgroup$
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
edited Dec 25 '18 at 17:12
Henning Makholm
241k17308549
asked Dec 25 '18 at 12:34
Maths SurvivorMaths Survivor
502219
edited Dec 25 '18 at 17:12
Henning Makholm
241k17308549
asked Dec 25 '18 at 12:34
Maths SurvivorMaths Survivor
502219
edited Dec 25 '18 at 17:12
Henning Makholm
241k17308549
edited Dec 25 '18 at 17:12
Henning Makholm
241k17308549
edited Dec 25 '18 at 17:12
Henning Makholm
241k17308549
241k17308549
asked Dec 25 '18 at 12:34
Maths SurvivorMaths Survivor
502219
asked Dec 25 '18 at 12:34
Maths SurvivorMaths Survivor
502219
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
|
show 3 more comments
$begingroup$
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
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Some comments:
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.
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).
You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.
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).
answered Dec 25 '18 at 13:05
Theo BenditTheo Bendit
19.4k12353
$endgroup$
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
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 17:11
Mark ViolaMark Viola
133k1277176
$endgroup$
$begingroup$
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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Some comments:
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.
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).
You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.
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).
answered Dec 25 '18 at 13:05
Theo BenditTheo Bendit
19.4k12353
$endgroup$
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
add a comment |
$begingroup$
Some comments:
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.
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).
You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.
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).
answered Dec 25 '18 at 13:05
Theo BenditTheo Bendit
19.4k12353
$endgroup$
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
add a comment |
$begingroup$
Some comments:
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.
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).
You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.
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).
answered Dec 25 '18 at 13:05
Theo BenditTheo Bendit
19.4k12353
$endgroup$
Some comments:
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.
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).
You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.
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).
answered Dec 25 '18 at 13:05
Theo BenditTheo Bendit
19.4k12353
answered Dec 25 '18 at 13:05
Theo BenditTheo Bendit
19.4k12353
answered Dec 25 '18 at 13:05
Theo BenditTheo Bendit
19.4k12353
answered Dec 25 '18 at 13:05
Theo BenditTheo Bendit
19.4k12353
19.4k12353
1
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"then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence?
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– Maths Survivor
Dec 25 '18 at 13:12
1
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@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
add a comment |
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
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 17:11
Mark ViolaMark Viola
133k1277176
$endgroup$
$begingroup$
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
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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
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 17:11
Mark ViolaMark Viola
133k1277176
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 17:11
Mark ViolaMark Viola
133k1277176
$endgroup$
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.
answered Dec 25 '18 at 17:11
Mark ViolaMark Viola
133k1277176
answered Dec 25 '18 at 17:11
Mark ViolaMark Viola
133k1277176
answered Dec 25 '18 at 17:11
Mark ViolaMark Viola
133k1277176
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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
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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