Hy, I am trying to solve this limit problem. First I see that this problem could be some integral sums problem but I couldn't see how soo please help me with explaining me this and giving me some other example.
$$\lim_{n\to\infty}\left(\ln\sqrt[n+1]{\frac{n+1}{n}}+\ln\sqrt[n+2]{\frac{n+2}{n}}+\cdots+\ln\sqrt[3n]{\frac{3n}{n}}\right)$$
This is the limit of a Riemann sum $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{2n}\frac{\ln\left(1+\frac{k}{n}\right)}{1+\frac{k}{n}}=\int_0^2 \frac{\ln\left(1+x\right)}{1+x}\,dx=\frac{1}{2}\left[\ln^2(1+x)\right]_0^2=\frac{\ln^2(3)}{2}.$$ More generally, for any integers $0<A<B$, $$\lim_{n\to\infty}\sum_{k=An+1}^{Bn}\ln\sqrt[k]{\frac{k}{n}}=\frac{\ln^2(B)-\ln^2(A)}{2}.$$
P.S. The given case is obtained from the general one for $B=3$ and $A=1$: $$\frac{1}{n}\sum_{k=1}^{2n}\frac{\ln\left(1+\frac{k}{n}\right)}{1+\frac{k}{n}}=\sum_{k=1}^{2n}\frac{\ln\left(\frac{n+k}{n}\right)}{n+k}=\sum_{j=n+1}^{3n}\frac{\ln\left(\frac{j}{n}\right)}{j} =\sum_{j=n+1}^{3n}\ln\sqrt[j]{\frac{j}{n}}$$ where $j=n+k$.
Keep in mind that if $f$ is continuous in $[0,1]$ then $\frac{1}{n}\sum_{k=1}^n f\left(\frac{k}{n}\right)\to \int_0^1f(x),dx$ and try to guess for the function $f$. See more examples:
http://math.stackexchange.com/questions/995058/the-limit-of-a-sum?rq=1
http://math.stackexchange.com/questions/1039221/expressing-limit-of-sum-definite-integral?rq=1
– Robert Z Aug 03 '16 at 14:53$$\lim_{n \rightarrow + \infty}\sum_{k=An+1}^{Bn}\ln\sqrt[k]{\frac{k}{n}}=\frac{\ln^2(B+1)-\ln^2(A+1)}{2}$$
– Tolaso Dec 15 '19 at 08:38
