How to find the lim when n goes to infinity of

Question

1) $$\sum_{k=1}^n \ln\left(1+\frac k{n^2}\right) =$$ so I tried solving it with Riemann but I don't know how to do it when I have n squared.(I was thinking of writing it as $$\int_{1}^b\ln(x) $$ .I found that b=2 (probably wrong)). 2) $$\sum_{k=1}^n \frac{1}{n+\sqrt{k+n^2}} =$$ I tried factoring 1/n and I found that the function would be $$\frac{1}{1+\sqrt{x^2+1}}$$

Answer 1

For any $k\in[1,n]$ we have $$ \log\left(1+\frac{k}{n^2}\right) = \frac{k}{n^2}-\frac{k^2}{2n^4}+O\left(\frac{k^3}{n^6}\right) $$ where $$ \lim_{n\to +\infty}\sum_{k=1}^{n}\frac{k}{n^2} = \int_{0}^{1}x\,dx = \frac{1}{2} $$ and $$ \lim_{n\to +\infty}\sum_{k=1}^{n}\left(\frac{k}{n^2}\right)^{m} = 0 $$ for any $m\geq 2$. In particular, the wanted limit is simply $\color{red}{\large\frac{1}{2}}$.

Answer 2

Hints:

For 1) use that for $x\geq 0$, you have $$\frac{x^2}{2}\geq x-\log(1+x)\geq 0$$(replace $x$ by $k/n^2$ and sum)

For 2) use that for $1\leq k\leq n$ you have $$2n\leq n+\sqrt{k+n^2}\leq n+\sqrt{n+n^2}$$