Problem Find the limit :
$\lim_{n\to \infty}\sqrt[n]{(1+1/n)(1+2/n)\cdot…\cdot(1+1/n)}$
which is same as problem solved here Evaluate the limit by first recognizing the sum as a Riemann Sum for a function defined on $[0,1]$., but in this case interval was given [0,1].
We apply the $\log$ function and we use the Riemann sum and we integrate:
$$\log\left(\frac{1}{n}\sqrt[n]\frac{(2n)!}{n!}\right)=-\log n+\frac{1}{n}\sum_{k=1}^n\log(k+n)=\frac{1}{n}\sum_{k=1}^n\log(k+n)-\log n\\=\frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}+1\right)\to\int_0^1 \log(1+x)dx=2\log(2)-1$$
hence
$$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]\frac{(2n)!}{n!}=\frac{4}{e}$$
Now if interval is not given as in my problem i get interval from (1,2) so i get different result.
if we follow that solution above we get $\frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}+1\right)$. Now from formula :
$\int_{a}^{b} f(x)dx=lim_{n\to\infty}\sum_{i=1}^{n}f(a+(\Delta x)i) $ i get a=1,b=2 interval so i get $\frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}+1\right)\to\int_1^2 \log(1+x)dx$.
My question is :
1.Is this correct bcs wolphram alpha gives https://www.wolframalpha.com/input/?i=limit+%281%2Fn%29*%28%282n%29%21%2Fn%21%29%5E%281%2Fn%29+as+n+goes+to+infinity%3De%5E%28integrate+log%281%2Bx%29dx%2C+x+goes+from+0+to+1%29 this is correct and this is not
https://www.wolframalpha.com/input/?i=limit+%281%2Fn%29*%28%282n%29%21%2Fn%21%29%5E%281%2Fn%29+as+n+goes+to+infinity%3De%5E%28integrate+log%281%2Bx%29dx%2C+x+goes+from+1+to+2%29
Best Answer
$\frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}+1\right)\to\int_1^2 \log(1+x)dx$ should be $\frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}+1\right)\to\int_0^1 \log(1+x)dx$.