Computation of: $\lim_{n\to\infty}\left(\ln\left(1+\frac{1}{n^2+1}\right)^n+\ldots+\ln\left(1+\frac{1}{n^2+n}\right)^n\right)$

Question

Evaluate: $$\lim_{n\to\infty}\left(\ln\left(1+\frac{1}{n^2+1}\right)^n+\ldots+\ln\left(1+\frac{1}{n^2+n}\right)^n\right)\;n\in\mathbb N$$

My attempt:

Using the manual limit: $$\lim_{x\to 0}\frac{\ln(1+x)}{x}=1$$

$$\lim_{n\to\infty}\left(\ln\left(1+\frac{1}{n^2+1}\right)^n+\ldots+\ln\left(1+\frac{1}{n^2+n}\right)^n\right)$$ $$=\lim_{n\to\infty}\left(\sum_{i=1}^n\frac{\ln\left(1+\frac{1}{n^2+i}\right)}{\frac{1}{n^2+i}}\cdot\frac{1}{n(n^2+i)}\right)=0$$

Is this correct?

Answer 1

$$\lim_{n\to\infty}\sum_{k=1}^{n}\ln\left(\left(1+\frac{1}{n^{2}+k}\right)^{n}\right)=\lim_{n\to\infty}n\ln\left(\prod_{k=1}^{n}\left(1+\frac{1}{n^{2}+k}\right)\right)=\lim_{n\to\infty}n\ln\left(\frac{n^{2}+n+1}{n^{2}+1}\right)=\color{red}{\boxed {1}}$$

Answer 2

You can squeeze it as follows:

$$n\sum_{k=1}^n\ln\left(1+\frac{1}{n^2+n}\right)\leq n\sum_{k=1}^n\ln\left(1+\frac{1}{n^2+k}\right)\leq n\sum_{k=1}^n\ln\left(1+\frac{1}{n^2}\right)$$

Hence, $$\underbrace{n^2\ln\left(1+\frac{1}{n^2+n}\right)}_{\stackrel{n\to\infty}{\longrightarrow}1}\leq n\sum_{k=1}^n\ln\left(1+\frac{1}{n^2+k}\right)\leq \underbrace{n^2\ln\left(1+\frac{1}{n^2}\right)}_{\stackrel{n\to\infty}{\longrightarrow}1}$$