value of Then $lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$
For a given sequence $a_{1},a_{2},a_{3},....,a_{n}$.
If $lim_{nrightarrow infty} a_{n} = a.$ Then $displaystyle lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$
Try: Using Stolz Method ,
Let $displaystyle frac{a_{n}}{b_{n}} = lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$
Then $$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} = lim_{nrightarrow infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})}=a_{n+1}$$
Answer given is $a$,
Could some help me Why $a_{n+1}=a_{n}$ for $nrightarrow infty,$ thanks
limits
asked Nov 29 at 14:22
D Tiwari
5,2952630
add a comment |
For a given sequence $a_{1},a_{2},a_{3},....,a_{n}$.
If $lim_{nrightarrow infty} a_{n} = a.$ Then $displaystyle lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$
Try: Using Stolz Method ,
Let $displaystyle frac{a_{n}}{b_{n}} = lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$
Then $$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} = lim_{nrightarrow infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})}=a_{n+1}$$
Answer given is $a$,
Could some help me Why $a_{n+1}=a_{n}$ for $nrightarrow infty,$ thanks
limits
asked Nov 29 at 14:22
D Tiwari
5,2952630
1
A more general result is true. If $a_nto L$ then any subsequence of $a_n$ (like $a_{n+1},a_{2n}$) also converges to $L$ and the proof is immediate via definition of limit. This however does not mean that $a_{n+1}=a_n$ for large $n$. It just means they have same limit. Two functions can have same limits without being equal.
– Paramanand Singh
Nov 30 at 2:58
add a comment |
For a given sequence $a_{1},a_{2},a_{3},....,a_{n}$.
If $lim_{nrightarrow infty} a_{n} = a.$ Then $displaystyle lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$
Try: Using Stolz Method ,
Let $displaystyle frac{a_{n}}{b_{n}} = lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$
Then $$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} = lim_{nrightarrow infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})}=a_{n+1}$$
Answer given is $a$,
Could some help me Why $a_{n+1}=a_{n}$ for $nrightarrow infty,$ thanks
limits
asked Nov 29 at 14:22
D Tiwari
5,2952630
For a given sequence $a_{1},a_{2},a_{3},....,a_{n}$.
If $lim_{nrightarrow infty} a_{n} = a.$ Then $displaystyle lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$
Try: Using Stolz Method ,
Let $displaystyle frac{a_{n}}{b_{n}} = lim_{nrightarrow infty}frac{1}{ln (n)}sum^{n}_{k=1}frac{a_{k}}{k}$
Then $$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} = lim_{nrightarrow infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})}=a_{n+1}$$
Answer given is $a$,
Could some help me Why $a_{n+1}=a_{n}$ for $nrightarrow infty,$ thanks
limits
limits
asked Nov 29 at 14:22
D Tiwari
5,2952630
asked Nov 29 at 14:22
D Tiwari
5,2952630
asked Nov 29 at 14:22
D Tiwari
5,2952630
asked Nov 29 at 14:22
D Tiwari
5,2952630
asked Nov 29 at 14:22
D Tiwari
5,2952630
5,2952630
1
A more general result is true. If $a_nto L$ then any subsequence of $a_n$ (like $a_{n+1},a_{2n}$) also converges to $L$ and the proof is immediate via definition of limit. This however does not mean that $a_{n+1}=a_n$ for large $n$. It just means they have same limit. Two functions can have same limits without being equal.
– Paramanand Singh
Nov 30 at 2:58
add a comment |
1
A more general result is true. If $a_nto L$ then any subsequence of $a_n$ (like $a_{n+1},a_{2n}$) also converges to $L$ and the proof is immediate via definition of limit. This however does not mean that $a_{n+1}=a_n$ for large $n$. It just means they have same limit. Two functions can have same limits without being equal.
– Paramanand Singh
Nov 30 at 2:58
1
1
A more general result is true. If $a_nto L$ then any subsequence of $a_n$ (like $a_{n+1},a_{2n}$) also converges to $L$ and the proof is immediate via definition of limit. This however does not mean that $a_{n+1}=a_n$ for large $n$. It just means they have same limit. Two functions can have same limits without being equal.
– Paramanand Singh
Nov 30 at 2:58
A more general result is true. If $a_nto L$ then any subsequence of $a_n$ (like $a_{n+1},a_{2n}$) also converges to $L$ and the proof is immediate via definition of limit. This however does not mean that $a_{n+1}=a_n$ for large $n$. It just means they have same limit. Two functions can have same limits without being equal.
– Paramanand Singh
Nov 30 at 2:58
add a comment |
2 Answers
2
active
oldest
votes
By Stolz-Cesaro
$$frac{a_{n}}{b_{n}} = lim_{nrightarrow infty} frac{sum^{n}_{k=1}frac{a_{k}}{k}}{ln (n)}$$
then
$$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} =lim_{nrightarrow infty}frac{frac{a_{n+1}}{n+1}}{ln (n+1)-ln n}=lim_{nrightarrow infty}frac{frac{n}{n+1}a_{n+1}}{ln left(1+frac1nright)^n} to a$$
answered Nov 29 at 14:58
gimusi
1
add a comment |
What the author of the solution is using is that $lim_{n to infty}a_{n+1} = lim_{n to infty}a_n$.
Therefore, we may rewrite the last step in a more rigorous way as
$lim_{n to infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})} = frac{lim_{n to infty}a_{n+1}}{lim_{n to infty}ln(1+frac{1}{n})^{n+1}} = frac{a}{log e} = a$, where I also used the fact that $ lim_{n to infty}(1+frac{1}{n})^{n+1} = e$
It doesn't have to be the case that $a_{n+1} = a_n$ for the solution to work.
answered Nov 29 at 14:36
Sorin Tirc
1,06511
I have a doubt why $lim_{nrightarrow infty}a_{n+1}=a.$ please explain me
– D Tiwari
Nov 29 at 14:52
@DurgeshTiwari Just by the definition of the limit: $lim_{n to infty}a_n = a$ is equivalent to $|a_n-a| < epsilon$ for all n suff large(for any fixed $epsilon$). This would also imply that $|a_{n+1}-a| < epsilon$ for all n sufficiently large and any (fixed) $epsilon$.
– Sorin Tirc
Nov 29 at 15:30
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
By Stolz-Cesaro
$$frac{a_{n}}{b_{n}} = lim_{nrightarrow infty} frac{sum^{n}_{k=1}frac{a_{k}}{k}}{ln (n)}$$
then
$$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} =lim_{nrightarrow infty}frac{frac{a_{n+1}}{n+1}}{ln (n+1)-ln n}=lim_{nrightarrow infty}frac{frac{n}{n+1}a_{n+1}}{ln left(1+frac1nright)^n} to a$$
answered Nov 29 at 14:58
gimusi
1
add a comment |
By Stolz-Cesaro
$$frac{a_{n}}{b_{n}} = lim_{nrightarrow infty} frac{sum^{n}_{k=1}frac{a_{k}}{k}}{ln (n)}$$
then
$$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} =lim_{nrightarrow infty}frac{frac{a_{n+1}}{n+1}}{ln (n+1)-ln n}=lim_{nrightarrow infty}frac{frac{n}{n+1}a_{n+1}}{ln left(1+frac1nright)^n} to a$$
answered Nov 29 at 14:58
gimusi
1
add a comment |
By Stolz-Cesaro
$$frac{a_{n}}{b_{n}} = lim_{nrightarrow infty} frac{sum^{n}_{k=1}frac{a_{k}}{k}}{ln (n)}$$
then
$$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} =lim_{nrightarrow infty}frac{frac{a_{n+1}}{n+1}}{ln (n+1)-ln n}=lim_{nrightarrow infty}frac{frac{n}{n+1}a_{n+1}}{ln left(1+frac1nright)^n} to a$$
answered Nov 29 at 14:58
gimusi
1
By Stolz-Cesaro
$$frac{a_{n}}{b_{n}} = lim_{nrightarrow infty} frac{sum^{n}_{k=1}frac{a_{k}}{k}}{ln (n)}$$
then
$$lim_{nrightarrow infty}frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} =lim_{nrightarrow infty}frac{frac{a_{n+1}}{n+1}}{ln (n+1)-ln n}=lim_{nrightarrow infty}frac{frac{n}{n+1}a_{n+1}}{ln left(1+frac1nright)^n} to a$$
answered Nov 29 at 14:58
gimusi
1
answered Nov 29 at 14:58
gimusi
1
answered Nov 29 at 14:58
gimusi
1
answered Nov 29 at 14:58
gimusi
1
1
add a comment |
add a comment |
What the author of the solution is using is that $lim_{n to infty}a_{n+1} = lim_{n to infty}a_n$.
Therefore, we may rewrite the last step in a more rigorous way as
$lim_{n to infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})} = frac{lim_{n to infty}a_{n+1}}{lim_{n to infty}ln(1+frac{1}{n})^{n+1}} = frac{a}{log e} = a$, where I also used the fact that $ lim_{n to infty}(1+frac{1}{n})^{n+1} = e$
It doesn't have to be the case that $a_{n+1} = a_n$ for the solution to work.
answered Nov 29 at 14:36
Sorin Tirc
1,06511
I have a doubt why $lim_{nrightarrow infty}a_{n+1}=a.$ please explain me
– D Tiwari
Nov 29 at 14:52
@DurgeshTiwari Just by the definition of the limit: $lim_{n to infty}a_n = a$ is equivalent to $|a_n-a| < epsilon$ for all n suff large(for any fixed $epsilon$). This would also imply that $|a_{n+1}-a| < epsilon$ for all n sufficiently large and any (fixed) $epsilon$.
– Sorin Tirc
Nov 29 at 15:30
add a comment |
What the author of the solution is using is that $lim_{n to infty}a_{n+1} = lim_{n to infty}a_n$.
Therefore, we may rewrite the last step in a more rigorous way as
$lim_{n to infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})} = frac{lim_{n to infty}a_{n+1}}{lim_{n to infty}ln(1+frac{1}{n})^{n+1}} = frac{a}{log e} = a$, where I also used the fact that $ lim_{n to infty}(1+frac{1}{n})^{n+1} = e$
It doesn't have to be the case that $a_{n+1} = a_n$ for the solution to work.
answered Nov 29 at 14:36
Sorin Tirc
1,06511
I have a doubt why $lim_{nrightarrow infty}a_{n+1}=a.$ please explain me
– D Tiwari
Nov 29 at 14:52
@DurgeshTiwari Just by the definition of the limit: $lim_{n to infty}a_n = a$ is equivalent to $|a_n-a| < epsilon$ for all n suff large(for any fixed $epsilon$). This would also imply that $|a_{n+1}-a| < epsilon$ for all n sufficiently large and any (fixed) $epsilon$.
– Sorin Tirc
Nov 29 at 15:30
add a comment |
What the author of the solution is using is that $lim_{n to infty}a_{n+1} = lim_{n to infty}a_n$.
Therefore, we may rewrite the last step in a more rigorous way as
$lim_{n to infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})} = frac{lim_{n to infty}a_{n+1}}{lim_{n to infty}ln(1+frac{1}{n})^{n+1}} = frac{a}{log e} = a$, where I also used the fact that $ lim_{n to infty}(1+frac{1}{n})^{n+1} = e$
It doesn't have to be the case that $a_{n+1} = a_n$ for the solution to work.
answered Nov 29 at 14:36
Sorin Tirc
1,06511
What the author of the solution is using is that $lim_{n to infty}a_{n+1} = lim_{n to infty}a_n$.
Therefore, we may rewrite the last step in a more rigorous way as
$lim_{n to infty}frac{a_{n+1}}{(n+1)ln(1+frac{1}{n})} = frac{lim_{n to infty}a_{n+1}}{lim_{n to infty}ln(1+frac{1}{n})^{n+1}} = frac{a}{log e} = a$, where I also used the fact that $ lim_{n to infty}(1+frac{1}{n})^{n+1} = e$
It doesn't have to be the case that $a_{n+1} = a_n$ for the solution to work.
answered Nov 29 at 14:36
Sorin Tirc
1,06511
answered Nov 29 at 14:36
Sorin Tirc
1,06511
answered Nov 29 at 14:36
Sorin Tirc
1,06511
answered Nov 29 at 14:36
Sorin Tirc
1,06511
1,06511
I have a doubt why $lim_{nrightarrow infty}a_{n+1}=a.$ please explain me
– D Tiwari
Nov 29 at 14:52
@DurgeshTiwari Just by the definition of the limit: $lim_{n to infty}a_n = a$ is equivalent to $|a_n-a| < epsilon$ for all n suff large(for any fixed $epsilon$). This would also imply that $|a_{n+1}-a| < epsilon$ for all n sufficiently large and any (fixed) $epsilon$.
– Sorin Tirc
Nov 29 at 15:30
add a comment |
I have a doubt why $lim_{nrightarrow infty}a_{n+1}=a.$ please explain me
– D Tiwari
Nov 29 at 14:52
@DurgeshTiwari Just by the definition of the limit: $lim_{n to infty}a_n = a$ is equivalent to $|a_n-a| < epsilon$ for all n suff large(for any fixed $epsilon$). This would also imply that $|a_{n+1}-a| < epsilon$ for all n sufficiently large and any (fixed) $epsilon$.
– Sorin Tirc
Nov 29 at 15:30
I have a doubt why $lim_{nrightarrow infty}a_{n+1}=a.$ please explain me
– D Tiwari
Nov 29 at 14:52
I have a doubt why $lim_{nrightarrow infty}a_{n+1}=a.$ please explain me
– D Tiwari
Nov 29 at 14:52
@DurgeshTiwari Just by the definition of the limit: $lim_{n to infty}a_n = a$ is equivalent to $|a_n-a| < epsilon$ for all n suff large(for any fixed $epsilon$). This would also imply that $|a_{n+1}-a| < epsilon$ for all n sufficiently large and any (fixed) $epsilon$.
– Sorin Tirc
Nov 29 at 15:30
@DurgeshTiwari Just by the definition of the limit: $lim_{n to infty}a_n = a$ is equivalent to $|a_n-a| < epsilon$ for all n suff large(for any fixed $epsilon$). This would also imply that $|a_{n+1}-a| < epsilon$ for all n sufficiently large and any (fixed) $epsilon$.
– Sorin Tirc
Nov 29 at 15:30
add a comment |
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1
A more general result is true. If $a_nto L$ then any subsequence of $a_n$ (like $a_{n+1},a_{2n}$) also converges to $L$ and the proof is immediate via definition of limit. This however does not mean that $a_{n+1}=a_n$ for large $n$. It just means they have same limit. Two functions can have same limits without being equal.
– Paramanand Singh
Nov 30 at 2:58