Compute limit of a sequence of integrals

Multi tool use
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0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 18:13









SvanN

2,0411422




2,0411422










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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,



$$ lim_{ntoinfty} int_1^2 f_n(x), mathrm{d}x = int_1^2 f(x), mathrm{d}x = int_1^2 x, mathrm{d}x = tfrac{3}{2}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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,



$$ lim_{ntoinfty} int_1^2 f_n(x), mathrm{d}x = int_1^2 f(x), mathrm{d}x = int_1^2 x, mathrm{d}x = tfrac{3}{2}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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,



$$ lim_{ntoinfty} int_1^2 f_n(x), mathrm{d}x = int_1^2 f(x), mathrm{d}x = int_1^2 x, mathrm{d}x = tfrac{3}{2}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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,



$$ lim_{ntoinfty} int_1^2 f_n(x), mathrm{d}x = int_1^2 f(x), mathrm{d}x = int_1^2 x, mathrm{d}x = tfrac{3}{2}.$$






share|cite|improve this answer









$endgroup$



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,



$$ lim_{ntoinfty} int_1^2 f_n(x), mathrm{d}x = int_1^2 f(x), mathrm{d}x = int_1^2 x, mathrm{d}x = tfrac{3}{2}.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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