Computing the limit of the sequence $\lim_{n\rightarrow\infty}\overset n{\underset{k=1}{\sum(}}\sqrt{1+\frac k{n^2}}-1)$ and another one

Question

I have to compute the limit of this sequence $\lim\limits_{n\to\infty}$ $\overset n{\underset{k=1}{\sum(}}\sqrt{1+\frac k{n^2}}-1)$. I tried to apply it's conjugate and then I go to this point: $\lim\limits_{n\rightarrow\infty}\overset n{\underset{k=1}{\sum(}}\frac k{n\sqrt{n^2+k}+n})$ and then tried to apply squeezing theorem but it gets me to a wrong answer...

The other sequence is: $\lim\limits_{n\rightarrow\infty}\frac1{3n+1}\sum\limits_{k=1}^n\cos\frac\pi{2n+k}$, for which I have no idea how to start it.

If you answer me can you provide me with some information about Limits of sequences and how to deal with them? Especially the ones with sums of square roots and trig sums.

Answer 1

Using $$1+\frac{x}{2}>\sqrt{1+x}>1+\frac{x}{2}-x^2$$ for $0<x<1$

Answer 2

For any $x\in(0,1)$ we have $\sqrt{1+x}-1=\frac{x}{2}+O(x^2)$, hence $$ \sum_{k=1}^{n}\sqrt{1+\frac{k}{n^2}}-1 =\frac{1}{2n^2}\sum_{k=1}^{n}k+O\left(\frac{1}{n^4}\sum_{k=1}^{n}k^2\right)=\color{red}{\frac{1}{4}}+O\left(\frac{1}{n}\right)$$ as $n\to +\infty$.