Compute limit of a sequence of integrals
$begingroup$
I cannot compute
$$lim_{ntoinfty}int_1^2left(1+frac{ln x}{n}right)^n ,mathrm{d}x.$$
The reason is that I cannot find an antiderivative for $(1+frac{ln x}{n})^n $,
or, I cannot see if the sequence ${(1+frac{ln x}{n})^n}_{n=1}^infty $ is uniformly convergent.
What should we do to find the limit in this case?
real-analysis analysis
edited Jan 17 at 18:13
SvanN
2,0411422
asked Jan 17 at 18:00
serenusserenus
24316
$endgroup$
add a comment |
$begingroup$
I cannot compute
$$lim_{ntoinfty}int_1^2left(1+frac{ln x}{n}right)^n ,mathrm{d}x.$$
The reason is that I cannot find an antiderivative for $(1+frac{ln x}{n})^n $,
or, I cannot see if the sequence ${(1+frac{ln x}{n})^n}_{n=1}^infty $ is uniformly convergent.
What should we do to find the limit in this case?
real-analysis analysis
edited Jan 17 at 18:13
SvanN
2,0411422
asked Jan 17 at 18:00
serenusserenus
24316
$endgroup$
$begingroup$
Are you allowed to use Lebesgue's dominated convergence theorem?
$endgroup$
– SvanN
Jan 17 at 18:03
$begingroup$
Yes, but what about the uniform convergence?
$endgroup$
– serenus
Jan 17 at 18:05
$begingroup$
If you use LDC, you do not need uniform convergence, only pointwise almost-everywhere.
$endgroup$
– SvanN
Jan 17 at 18:06
$begingroup$
Could you please post your answer?
$endgroup$
– serenus
Jan 17 at 18:07
add a comment |
$begingroup$
I cannot compute
$$lim_{ntoinfty}int_1^2left(1+frac{ln x}{n}right)^n ,mathrm{d}x.$$
The reason is that I cannot find an antiderivative for $(1+frac{ln x}{n})^n $,
or, I cannot see if the sequence ${(1+frac{ln x}{n})^n}_{n=1}^infty $ is uniformly convergent.
What should we do to find the limit in this case?
real-analysis analysis
edited Jan 17 at 18:13
SvanN
2,0411422
asked Jan 17 at 18:00
serenusserenus
24316
$endgroup$
I cannot compute
$$lim_{ntoinfty}int_1^2left(1+frac{ln x}{n}right)^n ,mathrm{d}x.$$
The reason is that I cannot find an antiderivative for $(1+frac{ln x}{n})^n $,
or, I cannot see if the sequence ${(1+frac{ln x}{n})^n}_{n=1}^infty $ is uniformly convergent.
What should we do to find the limit in this case?
real-analysis analysis
real-analysis analysis
edited Jan 17 at 18:13
SvanN
2,0411422
asked Jan 17 at 18:00
serenusserenus
24316
edited Jan 17 at 18:13
SvanN
2,0411422
asked Jan 17 at 18:00
serenusserenus
24316
edited Jan 17 at 18:13
SvanN
2,0411422
edited Jan 17 at 18:13
SvanN
2,0411422
edited Jan 17 at 18:13
SvanN
2,0411422
2,0411422
asked Jan 17 at 18:00
serenusserenus
24316
asked Jan 17 at 18:00
serenusserenus
24316
asked Jan 17 at 18:00
serenusserenus
24316
24316
$begingroup$
Are you allowed to use Lebesgue's dominated convergence theorem?
$endgroup$
– SvanN
Jan 17 at 18:03
$begingroup$
Yes, but what about the uniform convergence?
$endgroup$
– serenus
Jan 17 at 18:05
$begingroup$
If you use LDC, you do not need uniform convergence, only pointwise almost-everywhere.
$endgroup$
– SvanN
Jan 17 at 18:06
$begingroup$
Could you please post your answer?
$endgroup$
– serenus
Jan 17 at 18:07
add a comment |
$begingroup$
Are you allowed to use Lebesgue's dominated convergence theorem?
$endgroup$
– SvanN
Jan 17 at 18:03
$begingroup$
Yes, but what about the uniform convergence?
$endgroup$
– serenus
Jan 17 at 18:05
$begingroup$
If you use LDC, you do not need uniform convergence, only pointwise almost-everywhere.
$endgroup$
– SvanN
Jan 17 at 18:06
$begingroup$
Could you please post your answer?
$endgroup$
– serenus
Jan 17 at 18:07
$begingroup$
Are you allowed to use Lebesgue's dominated convergence theorem?
$endgroup$
– SvanN
Jan 17 at 18:03
$begingroup$
Are you allowed to use Lebesgue's dominated convergence theorem?
$endgroup$
– SvanN
Jan 17 at 18:03
$begingroup$
Yes, but what about the uniform convergence?
$endgroup$
– serenus
Jan 17 at 18:05
$begingroup$
Yes, but what about the uniform convergence?
$endgroup$
– serenus
Jan 17 at 18:05
$begingroup$
If you use LDC, you do not need uniform convergence, only pointwise almost-everywhere.
$endgroup$
– SvanN
Jan 17 at 18:06
$begingroup$
If you use LDC, you do not need uniform convergence, only pointwise almost-everywhere.
$endgroup$
– SvanN
Jan 17 at 18:06
$begingroup$
Could you please post your answer?
$endgroup$
– serenus
Jan 17 at 18:07
$begingroup$
Could you please post your answer?
$endgroup$
– serenus
Jan 17 at 18:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First, note that for every $x in [1,2]$,
$$ lim_{ntoinfty} left(1+frac{ln x}{n}right)^n = e^{ln x} =x.$$
This limit is increasing, so that all the functions $f_n(x) := left(1+frac{ln x}{n}right)^n$ are bounded from above by $f(x) := x$ on $[1,2]$.
We also have pointwise convergence $f_n to f$ on $[1,2]$, so that by Lebesgue's dominated convergence theorem,
$$ lim_{ntoinfty} int_1^2 f_n(x), mathrm{d}x = int_1^2 f(x), mathrm{d}x = int_1^2 x, mathrm{d}x = tfrac{3}{2}.$$
answered Jan 17 at 18:09
SvanNSvanN
2,0411422
$endgroup$
$begingroup$
Thanks for the answer.
$endgroup$
– serenus
Jan 17 at 18:16
$begingroup$
I think we do not need even to know that the sequence ${(1+frac{lnx}{n})^n }_{n=1}^infty$ is increasing, because note that on $[1,2]$, $lnxleq lne=1$, so $(1+frac{lnx}{n})^nleq (1+frac{1}{n})^n leq e$ and we can take the function $f$ as $f(x):=e$ on $[1,2]$. But, of course we should know that the sequence ${(1+frac{1}{n})^n }_{n=1}^infty$ is increasing.
$endgroup$
– serenus
Jan 18 at 10:40
$begingroup$
@serenus You're still using that $(1+1/n)^n$ is increasing in $n$ if you want to bound it by $e$ for all $n$. Though indeed it is not strictly necessary that it be increasing; the functions $f_n$ just need to be bounded by some integrable function on $[1,2]$. The fact that the limit is increasing just made the argument even simpler: we can bound them by their (pointwise) limit.
$endgroup$
– SvanN
Jan 18 at 10:42
$begingroup$
Right, I agree with you.
$endgroup$
– serenus
Jan 18 at 10:44
add a comment |
Your Answer
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1 Answer
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active
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1 Answer
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oldest
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oldest
votes
$begingroup$
First, note that for every $x in [1,2]$,
$$ lim_{ntoinfty} left(1+frac{ln x}{n}right)^n = e^{ln x} =x.$$
This limit is increasing, so that all the functions $f_n(x) := left(1+frac{ln x}{n}right)^n$ are bounded from above by $f(x) := x$ on $[1,2]$.
We also have pointwise convergence $f_n to f$ on $[1,2]$, so that by Lebesgue's dominated convergence theorem,
$$ lim_{ntoinfty} int_1^2 f_n(x), mathrm{d}x = int_1^2 f(x), mathrm{d}x = int_1^2 x, mathrm{d}x = tfrac{3}{2}.$$
answered Jan 17 at 18:09
SvanNSvanN
2,0411422
$endgroup$
$begingroup$
Thanks for the answer.
$endgroup$
– serenus
Jan 17 at 18:16
$begingroup$
I think we do not need even to know that the sequence ${(1+frac{lnx}{n})^n }_{n=1}^infty$ is increasing, because note that on $[1,2]$, $lnxleq lne=1$, so $(1+frac{lnx}{n})^nleq (1+frac{1}{n})^n leq e$ and we can take the function $f$ as $f(x):=e$ on $[1,2]$. But, of course we should know that the sequence ${(1+frac{1}{n})^n }_{n=1}^infty$ is increasing.
$endgroup$
– serenus
Jan 18 at 10:40
$begingroup$
@serenus You're still using that $(1+1/n)^n$ is increasing in $n$ if you want to bound it by $e$ for all $n$. Though indeed it is not strictly necessary that it be increasing; the functions $f_n$ just need to be bounded by some integrable function on $[1,2]$. The fact that the limit is increasing just made the argument even simpler: we can bound them by their (pointwise) limit.
$endgroup$
– SvanN
Jan 18 at 10:42
$begingroup$
Right, I agree with you.
$endgroup$
– serenus
Jan 18 at 10:44
add a comment |
$begingroup$
First, note that for every $x in [1,2]$,
$$ lim_{ntoinfty} left(1+frac{ln x}{n}right)^n = e^{ln x} =x.$$
This limit is increasing, so that all the functions $f_n(x) := left(1+frac{ln x}{n}right)^n$ are bounded from above by $f(x) := x$ on $[1,2]$.
We also have pointwise convergence $f_n to f$ on $[1,2]$, so that by Lebesgue's dominated convergence theorem,
$$ lim_{ntoinfty} int_1^2 f_n(x), mathrm{d}x = int_1^2 f(x), mathrm{d}x = int_1^2 x, mathrm{d}x = tfrac{3}{2}.$$
answered Jan 17 at 18:09
SvanNSvanN
2,0411422
$endgroup$
$begingroup$
Thanks for the answer.
$endgroup$
– serenus
Jan 17 at 18:16
$begingroup$
I think we do not need even to know that the sequence ${(1+frac{lnx}{n})^n }_{n=1}^infty$ is increasing, because note that on $[1,2]$, $lnxleq lne=1$, so $(1+frac{lnx}{n})^nleq (1+frac{1}{n})^n leq e$ and we can take the function $f$ as $f(x):=e$ on $[1,2]$. But, of course we should know that the sequence ${(1+frac{1}{n})^n }_{n=1}^infty$ is increasing.
$endgroup$
– serenus
Jan 18 at 10:40
$begingroup$
@serenus You're still using that $(1+1/n)^n$ is increasing in $n$ if you want to bound it by $e$ for all $n$. Though indeed it is not strictly necessary that it be increasing; the functions $f_n$ just need to be bounded by some integrable function on $[1,2]$. The fact that the limit is increasing just made the argument even simpler: we can bound them by their (pointwise) limit.
$endgroup$
– SvanN
Jan 18 at 10:42
$begingroup$
Right, I agree with you.
$endgroup$
– serenus
Jan 18 at 10:44
add a comment |
$begingroup$
First, note that for every $x in [1,2]$,
$$ lim_{ntoinfty} left(1+frac{ln x}{n}right)^n = e^{ln x} =x.$$
This limit is increasing, so that all the functions $f_n(x) := left(1+frac{ln x}{n}right)^n$ are bounded from above by $f(x) := x$ on $[1,2]$.
We also have pointwise convergence $f_n to f$ on $[1,2]$, so that by Lebesgue's dominated convergence theorem,
$$ lim_{ntoinfty} int_1^2 f_n(x), mathrm{d}x = int_1^2 f(x), mathrm{d}x = int_1^2 x, mathrm{d}x = tfrac{3}{2}.$$
answered Jan 17 at 18:09
SvanNSvanN
2,0411422
$endgroup$
First, note that for every $x in [1,2]$,
$$ lim_{ntoinfty} left(1+frac{ln x}{n}right)^n = e^{ln x} =x.$$
This limit is increasing, so that all the functions $f_n(x) := left(1+frac{ln x}{n}right)^n$ are bounded from above by $f(x) := x$ on $[1,2]$.
We also have pointwise convergence $f_n to f$ on $[1,2]$, so that by Lebesgue's dominated convergence theorem,
$$ lim_{ntoinfty} int_1^2 f_n(x), mathrm{d}x = int_1^2 f(x), mathrm{d}x = int_1^2 x, mathrm{d}x = tfrac{3}{2}.$$
answered Jan 17 at 18:09
SvanNSvanN
2,0411422
answered Jan 17 at 18:09
SvanNSvanN
2,0411422
answered Jan 17 at 18:09
SvanNSvanN
2,0411422
answered Jan 17 at 18:09
SvanNSvanN
2,0411422
2,0411422
$begingroup$
Thanks for the answer.
$endgroup$
– serenus
Jan 17 at 18:16
$begingroup$
I think we do not need even to know that the sequence ${(1+frac{lnx}{n})^n }_{n=1}^infty$ is increasing, because note that on $[1,2]$, $lnxleq lne=1$, so $(1+frac{lnx}{n})^nleq (1+frac{1}{n})^n leq e$ and we can take the function $f$ as $f(x):=e$ on $[1,2]$. But, of course we should know that the sequence ${(1+frac{1}{n})^n }_{n=1}^infty$ is increasing.
$endgroup$
– serenus
Jan 18 at 10:40
$begingroup$
@serenus You're still using that $(1+1/n)^n$ is increasing in $n$ if you want to bound it by $e$ for all $n$. Though indeed it is not strictly necessary that it be increasing; the functions $f_n$ just need to be bounded by some integrable function on $[1,2]$. The fact that the limit is increasing just made the argument even simpler: we can bound them by their (pointwise) limit.
$endgroup$
– SvanN
Jan 18 at 10:42
$begingroup$
Right, I agree with you.
$endgroup$
– serenus
Jan 18 at 10:44
add a comment |
$begingroup$
Thanks for the answer.
$endgroup$
– serenus
Jan 17 at 18:16
$begingroup$
I think we do not need even to know that the sequence ${(1+frac{lnx}{n})^n }_{n=1}^infty$ is increasing, because note that on $[1,2]$, $lnxleq lne=1$, so $(1+frac{lnx}{n})^nleq (1+frac{1}{n})^n leq e$ and we can take the function $f$ as $f(x):=e$ on $[1,2]$. But, of course we should know that the sequence ${(1+frac{1}{n})^n }_{n=1}^infty$ is increasing.
$endgroup$
– serenus
Jan 18 at 10:40
$begingroup$
@serenus You're still using that $(1+1/n)^n$ is increasing in $n$ if you want to bound it by $e$ for all $n$. Though indeed it is not strictly necessary that it be increasing; the functions $f_n$ just need to be bounded by some integrable function on $[1,2]$. The fact that the limit is increasing just made the argument even simpler: we can bound them by their (pointwise) limit.
$endgroup$
– SvanN
Jan 18 at 10:42
$begingroup$
Right, I agree with you.
$endgroup$
– serenus
Jan 18 at 10:44
$begingroup$
Thanks for the answer.
$endgroup$
– serenus
Jan 17 at 18:16
$begingroup$
Thanks for the answer.
$endgroup$
– serenus
Jan 17 at 18:16
$begingroup$
I think we do not need even to know that the sequence ${(1+frac{lnx}{n})^n }_{n=1}^infty$ is increasing, because note that on $[1,2]$, $lnxleq lne=1$, so $(1+frac{lnx}{n})^nleq (1+frac{1}{n})^n leq e$ and we can take the function $f$ as $f(x):=e$ on $[1,2]$. But, of course we should know that the sequence ${(1+frac{1}{n})^n }_{n=1}^infty$ is increasing.
$endgroup$
– serenus
Jan 18 at 10:40
$begingroup$
I think we do not need even to know that the sequence ${(1+frac{lnx}{n})^n }_{n=1}^infty$ is increasing, because note that on $[1,2]$, $lnxleq lne=1$, so $(1+frac{lnx}{n})^nleq (1+frac{1}{n})^n leq e$ and we can take the function $f$ as $f(x):=e$ on $[1,2]$. But, of course we should know that the sequence ${(1+frac{1}{n})^n }_{n=1}^infty$ is increasing.
$endgroup$
– serenus
Jan 18 at 10:40
$begingroup$
@serenus You're still using that $(1+1/n)^n$ is increasing in $n$ if you want to bound it by $e$ for all $n$. Though indeed it is not strictly necessary that it be increasing; the functions $f_n$ just need to be bounded by some integrable function on $[1,2]$. The fact that the limit is increasing just made the argument even simpler: we can bound them by their (pointwise) limit.
$endgroup$
– SvanN
Jan 18 at 10:42
$begingroup$
@serenus You're still using that $(1+1/n)^n$ is increasing in $n$ if you want to bound it by $e$ for all $n$. Though indeed it is not strictly necessary that it be increasing; the functions $f_n$ just need to be bounded by some integrable function on $[1,2]$. The fact that the limit is increasing just made the argument even simpler: we can bound them by their (pointwise) limit.
$endgroup$
– SvanN
Jan 18 at 10:42
$begingroup$
Right, I agree with you.
$endgroup$
– serenus
Jan 18 at 10:44
$begingroup$
Right, I agree with you.
$endgroup$
– serenus
Jan 18 at 10:44
add a comment |
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$begingroup$
Are you allowed to use Lebesgue's dominated convergence theorem?
$endgroup$
– SvanN
Jan 17 at 18:03
$begingroup$
Yes, but what about the uniform convergence?
$endgroup$
– serenus
Jan 17 at 18:05
$begingroup$
If you use LDC, you do not need uniform convergence, only pointwise almost-everywhere.
$endgroup$
– SvanN
Jan 17 at 18:06
$begingroup$
Could you please post your answer?
$endgroup$
– serenus
Jan 17 at 18:07