To evaluate the limit.
up vote
-1
down vote
favorite
Evaluate
$$lim_{ntoinfty}left[left(frac1nright)^n+left(frac2nright)^n+cdots+left(frac{n-1}nright)^n+left(frac nnright)^nright]$$
I tried to solve this by taking logs on both sides and got this form:
$$log l = lim_{ntoinfty} nlogfrac{n!}{n^n}=lim_{ntoinfty}nlogfrac{n!}{n^n}$$
After applying limit, $frac{n!}{n^n}to0$, so $logfrac{n!}{n^n}to-infty$ as $ntoinfty$. Then I am stuck. How to solve these type of problems?
limits
edited Nov 18 at 4:34
Rócherz
2,6612721
asked Nov 18 at 3:03
Mathsaddict
607
|
show 1 more comment
up vote
-1
down vote
favorite
Evaluate
$$lim_{ntoinfty}left[left(frac1nright)^n+left(frac2nright)^n+cdots+left(frac{n-1}nright)^n+left(frac nnright)^nright]$$
I tried to solve this by taking logs on both sides and got this form:
$$log l = lim_{ntoinfty} nlogfrac{n!}{n^n}=lim_{ntoinfty}nlogfrac{n!}{n^n}$$
After applying limit, $frac{n!}{n^n}to0$, so $logfrac{n!}{n^n}to-infty$ as $ntoinfty$. Then I am stuck. How to solve these type of problems?
limits
edited Nov 18 at 4:34
Rócherz
2,6612721
asked Nov 18 at 3:03
Mathsaddict
607
Better formatting please. Also, how did taking log of a sum convert it into a product?
– Aditya Dua
Nov 18 at 3:10
Yes, I made mistake.
– Mathsaddict
Nov 18 at 3:16
@Mathsaddict: I just edit your question for a good look. Is I'm missing anything?
– Chinnapparaj R
Nov 18 at 3:19
@Chinnapparaj you didn't miss anything. It looks better now. Thanks for edit.
– Mathsaddict
Nov 18 at 3:22
1
It is $lim_{n→∞}frac{1}{n^2}.frac{n(n+1)(2n+1)}{6}=∞$
– sirous
Nov 18 at 3:56
|
show 1 more comment
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Evaluate
$$lim_{ntoinfty}left[left(frac1nright)^n+left(frac2nright)^n+cdots+left(frac{n-1}nright)^n+left(frac nnright)^nright]$$
I tried to solve this by taking logs on both sides and got this form:
$$log l = lim_{ntoinfty} nlogfrac{n!}{n^n}=lim_{ntoinfty}nlogfrac{n!}{n^n}$$
After applying limit, $frac{n!}{n^n}to0$, so $logfrac{n!}{n^n}to-infty$ as $ntoinfty$. Then I am stuck. How to solve these type of problems?
limits
edited Nov 18 at 4:34
Rócherz
2,6612721
asked Nov 18 at 3:03
Mathsaddict
607
Evaluate
$$lim_{ntoinfty}left[left(frac1nright)^n+left(frac2nright)^n+cdots+left(frac{n-1}nright)^n+left(frac nnright)^nright]$$
I tried to solve this by taking logs on both sides and got this form:
$$log l = lim_{ntoinfty} nlogfrac{n!}{n^n}=lim_{ntoinfty}nlogfrac{n!}{n^n}$$
After applying limit, $frac{n!}{n^n}to0$, so $logfrac{n!}{n^n}to-infty$ as $ntoinfty$. Then I am stuck. How to solve these type of problems?
limits
limits
edited Nov 18 at 4:34
Rócherz
2,6612721
asked Nov 18 at 3:03
Mathsaddict
607
edited Nov 18 at 4:34
Rócherz
2,6612721
asked Nov 18 at 3:03
Mathsaddict
607
edited Nov 18 at 4:34
Rócherz
2,6612721
edited Nov 18 at 4:34
Rócherz
2,6612721
edited Nov 18 at 4:34
Rócherz
2,6612721
2,6612721
asked Nov 18 at 3:03
Mathsaddict
607
asked Nov 18 at 3:03
Mathsaddict
607
asked Nov 18 at 3:03
Mathsaddict
607
607
Better formatting please. Also, how did taking log of a sum convert it into a product?
– Aditya Dua
Nov 18 at 3:10
Yes, I made mistake.
– Mathsaddict
Nov 18 at 3:16
@Mathsaddict: I just edit your question for a good look. Is I'm missing anything?
– Chinnapparaj R
Nov 18 at 3:19
@Chinnapparaj you didn't miss anything. It looks better now. Thanks for edit.
– Mathsaddict
Nov 18 at 3:22
1
It is $lim_{n→∞}frac{1}{n^2}.frac{n(n+1)(2n+1)}{6}=∞$
– sirous
Nov 18 at 3:56
|
show 1 more comment
Better formatting please. Also, how did taking log of a sum convert it into a product?
– Aditya Dua
Nov 18 at 3:10
Yes, I made mistake.
– Mathsaddict
Nov 18 at 3:16
@Mathsaddict: I just edit your question for a good look. Is I'm missing anything?
– Chinnapparaj R
Nov 18 at 3:19
@Chinnapparaj you didn't miss anything. It looks better now. Thanks for edit.
– Mathsaddict
Nov 18 at 3:22
1
It is $lim_{n→∞}frac{1}{n^2}.frac{n(n+1)(2n+1)}{6}=∞$
– sirous
Nov 18 at 3:56
Better formatting please. Also, how did taking log of a sum convert it into a product?
– Aditya Dua
Nov 18 at 3:10
Better formatting please. Also, how did taking log of a sum convert it into a product?
– Aditya Dua
Nov 18 at 3:10
Yes, I made mistake.
– Mathsaddict
Nov 18 at 3:16
Yes, I made mistake.
– Mathsaddict
Nov 18 at 3:16
@Mathsaddict: I just edit your question for a good look. Is I'm missing anything?
– Chinnapparaj R
Nov 18 at 3:19
@Mathsaddict: I just edit your question for a good look. Is I'm missing anything?
– Chinnapparaj R
Nov 18 at 3:19
@Chinnapparaj you didn't miss anything. It looks better now. Thanks for edit.
– Mathsaddict
Nov 18 at 3:22
@Chinnapparaj you didn't miss anything. It looks better now. Thanks for edit.
– Mathsaddict
Nov 18 at 3:22
1
1
It is $lim_{n→∞}frac{1}{n^2}.frac{n(n+1)(2n+1)}{6}=∞$
– sirous
Nov 18 at 3:56
It is $lim_{n→∞}frac{1}{n^2}.frac{n(n+1)(2n+1)}{6}=∞$
– sirous
Nov 18 at 3:56
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Correct version of the question:
$$lim_{ntoinfty}sum_{k=1}^{n}left(frac{k}{n}right)^n=? $$
And this sum converges to
$$
lim_{ntoinfty}sum_{k=0}^{infty}left(frac{n-k}{n}right)^n1_{kleq n} = sum_{k=0}^infty lim_{ntoinfty}left(frac{n-k}{n}right)^n1_{kleq n} = sum_{k=0}^infty e^{-k} = frac{1}{1-e^{-1}},
$$ by (Lebesgue's) dominated convergence theorem or monotone convergence theorem. (Note: $(1-frac{k}{n})^n uparrow e^{-k}$ as $nto infty$.)
answered Nov 18 at 4:36
Song
82011
I didn't understand the transformation from summation(k=1 to n) (k/n)^n to summation(k=0 to ∞) ((n-k)/n)^n
– Mathsaddict
Nov 18 at 4:53
Sorry that I didn't explain the notation $1_{kleq n}.$ This 'indicator' function takes value $1$ if $kleq n$ and $0$ otherwise, so infinite sum is actually calculated only for $0leq kleq n$.
– Song
Nov 18 at 4:57
Okay, I got it. Thanks.
– Mathsaddict
Nov 18 at 5:04
1
There can be possibly so many different ways to get the limit result of the given sum, but one of the things that you should do first is to investigate the behavior of each term, guess how the total sum would behave and choose/find the appropriate tools/theorems that can show your intuition rigorously.
– Song
Nov 18 at 9:30
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Correct version of the question:
$$lim_{ntoinfty}sum_{k=1}^{n}left(frac{k}{n}right)^n=? $$
And this sum converges to
$$
lim_{ntoinfty}sum_{k=0}^{infty}left(frac{n-k}{n}right)^n1_{kleq n} = sum_{k=0}^infty lim_{ntoinfty}left(frac{n-k}{n}right)^n1_{kleq n} = sum_{k=0}^infty e^{-k} = frac{1}{1-e^{-1}},
$$ by (Lebesgue's) dominated convergence theorem or monotone convergence theorem. (Note: $(1-frac{k}{n})^n uparrow e^{-k}$ as $nto infty$.)
answered Nov 18 at 4:36
Song
82011
I didn't understand the transformation from summation(k=1 to n) (k/n)^n to summation(k=0 to ∞) ((n-k)/n)^n
– Mathsaddict
Nov 18 at 4:53
Sorry that I didn't explain the notation $1_{kleq n}.$ This 'indicator' function takes value $1$ if $kleq n$ and $0$ otherwise, so infinite sum is actually calculated only for $0leq kleq n$.
– Song
Nov 18 at 4:57
Okay, I got it. Thanks.
– Mathsaddict
Nov 18 at 5:04
1
There can be possibly so many different ways to get the limit result of the given sum, but one of the things that you should do first is to investigate the behavior of each term, guess how the total sum would behave and choose/find the appropriate tools/theorems that can show your intuition rigorously.
– Song
Nov 18 at 9:30
add a comment |
up vote
4
down vote
accepted
Correct version of the question:
$$lim_{ntoinfty}sum_{k=1}^{n}left(frac{k}{n}right)^n=? $$
And this sum converges to
$$
lim_{ntoinfty}sum_{k=0}^{infty}left(frac{n-k}{n}right)^n1_{kleq n} = sum_{k=0}^infty lim_{ntoinfty}left(frac{n-k}{n}right)^n1_{kleq n} = sum_{k=0}^infty e^{-k} = frac{1}{1-e^{-1}},
$$ by (Lebesgue's) dominated convergence theorem or monotone convergence theorem. (Note: $(1-frac{k}{n})^n uparrow e^{-k}$ as $nto infty$.)
answered Nov 18 at 4:36
Song
82011
I didn't understand the transformation from summation(k=1 to n) (k/n)^n to summation(k=0 to ∞) ((n-k)/n)^n
– Mathsaddict
Nov 18 at 4:53
Sorry that I didn't explain the notation $1_{kleq n}.$ This 'indicator' function takes value $1$ if $kleq n$ and $0$ otherwise, so infinite sum is actually calculated only for $0leq kleq n$.
– Song
Nov 18 at 4:57
Okay, I got it. Thanks.
– Mathsaddict
Nov 18 at 5:04
1
There can be possibly so many different ways to get the limit result of the given sum, but one of the things that you should do first is to investigate the behavior of each term, guess how the total sum would behave and choose/find the appropriate tools/theorems that can show your intuition rigorously.
– Song
Nov 18 at 9:30
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Correct version of the question:
$$lim_{ntoinfty}sum_{k=1}^{n}left(frac{k}{n}right)^n=? $$
And this sum converges to
$$
lim_{ntoinfty}sum_{k=0}^{infty}left(frac{n-k}{n}right)^n1_{kleq n} = sum_{k=0}^infty lim_{ntoinfty}left(frac{n-k}{n}right)^n1_{kleq n} = sum_{k=0}^infty e^{-k} = frac{1}{1-e^{-1}},
$$ by (Lebesgue's) dominated convergence theorem or monotone convergence theorem. (Note: $(1-frac{k}{n})^n uparrow e^{-k}$ as $nto infty$.)
answered Nov 18 at 4:36
Song
82011
Correct version of the question:
$$lim_{ntoinfty}sum_{k=1}^{n}left(frac{k}{n}right)^n=? $$
And this sum converges to
$$
lim_{ntoinfty}sum_{k=0}^{infty}left(frac{n-k}{n}right)^n1_{kleq n} = sum_{k=0}^infty lim_{ntoinfty}left(frac{n-k}{n}right)^n1_{kleq n} = sum_{k=0}^infty e^{-k} = frac{1}{1-e^{-1}},
$$ by (Lebesgue's) dominated convergence theorem or monotone convergence theorem. (Note: $(1-frac{k}{n})^n uparrow e^{-k}$ as $nto infty$.)
answered Nov 18 at 4:36
Song
82011
edited Nov 18 at 4:44
answered Nov 18 at 4:36
Song
82011
answered Nov 18 at 4:36
Song
82011
answered Nov 18 at 4:36
Song
82011
82011
I didn't understand the transformation from summation(k=1 to n) (k/n)^n to summation(k=0 to ∞) ((n-k)/n)^n
– Mathsaddict
Nov 18 at 4:53
Sorry that I didn't explain the notation $1_{kleq n}.$ This 'indicator' function takes value $1$ if $kleq n$ and $0$ otherwise, so infinite sum is actually calculated only for $0leq kleq n$.
– Song
Nov 18 at 4:57
Okay, I got it. Thanks.
– Mathsaddict
Nov 18 at 5:04
1
There can be possibly so many different ways to get the limit result of the given sum, but one of the things that you should do first is to investigate the behavior of each term, guess how the total sum would behave and choose/find the appropriate tools/theorems that can show your intuition rigorously.
– Song
Nov 18 at 9:30
add a comment |
I didn't understand the transformation from summation(k=1 to n) (k/n)^n to summation(k=0 to ∞) ((n-k)/n)^n
– Mathsaddict
Nov 18 at 4:53
Sorry that I didn't explain the notation $1_{kleq n}.$ This 'indicator' function takes value $1$ if $kleq n$ and $0$ otherwise, so infinite sum is actually calculated only for $0leq kleq n$.
– Song
Nov 18 at 4:57
Okay, I got it. Thanks.
– Mathsaddict
Nov 18 at 5:04
1
There can be possibly so many different ways to get the limit result of the given sum, but one of the things that you should do first is to investigate the behavior of each term, guess how the total sum would behave and choose/find the appropriate tools/theorems that can show your intuition rigorously.
– Song
Nov 18 at 9:30
I didn't understand the transformation from summation(k=1 to n) (k/n)^n to summation(k=0 to ∞) ((n-k)/n)^n
– Mathsaddict
Nov 18 at 4:53
I didn't understand the transformation from summation(k=1 to n) (k/n)^n to summation(k=0 to ∞) ((n-k)/n)^n
– Mathsaddict
Nov 18 at 4:53
Sorry that I didn't explain the notation $1_{kleq n}.$ This 'indicator' function takes value $1$ if $kleq n$ and $0$ otherwise, so infinite sum is actually calculated only for $0leq kleq n$.
– Song
Nov 18 at 4:57
Sorry that I didn't explain the notation $1_{kleq n}.$ This 'indicator' function takes value $1$ if $kleq n$ and $0$ otherwise, so infinite sum is actually calculated only for $0leq kleq n$.
– Song
Nov 18 at 4:57
Okay, I got it. Thanks.
– Mathsaddict
Nov 18 at 5:04
Okay, I got it. Thanks.
– Mathsaddict
Nov 18 at 5:04
1
1
There can be possibly so many different ways to get the limit result of the given sum, but one of the things that you should do first is to investigate the behavior of each term, guess how the total sum would behave and choose/find the appropriate tools/theorems that can show your intuition rigorously.
– Song
Nov 18 at 9:30
There can be possibly so many different ways to get the limit result of the given sum, but one of the things that you should do first is to investigate the behavior of each term, guess how the total sum would behave and choose/find the appropriate tools/theorems that can show your intuition rigorously.
– Song
Nov 18 at 9:30
add a comment |
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Better formatting please. Also, how did taking log of a sum convert it into a product?
– Aditya Dua
Nov 18 at 3:10
Yes, I made mistake.
– Mathsaddict
Nov 18 at 3:16
@Mathsaddict: I just edit your question for a good look. Is I'm missing anything?
– Chinnapparaj R
Nov 18 at 3:19
@Chinnapparaj you didn't miss anything. It looks better now. Thanks for edit.
– Mathsaddict
Nov 18 at 3:22
1
It is $lim_{n→∞}frac{1}{n^2}.frac{n(n+1)(2n+1)}{6}=∞$
– sirous
Nov 18 at 3:56