study of two sequences I need to study wether thos two sequences converge or not.
1) $u_n=n\sum_{k=1}^{2n+1} \frac{1}{n^2+k}$
2) $v_n=\frac{1}{n}\sum_{k=0}^{n-1} \cos(\frac{1}{\sqrt{n+k}})$
For the first i get it converges to $0$ by just expressing the sum as a fraction (not very smart imo).
 A: *

*Note that $(n^{2} +k) \leq (n^{2}+2n+1)$ whenever $k \in\left\{1,2,\cdots,(2n+1)\right\}$. So we have that $\frac{1}{n^{2}+k} \geq \frac{1}{n^{2}+2n+1} =\frac{1}{(n+1)^{2}}$. Similarly $n^{2}+1 \leq n^{2}+k$ and so we have that $\frac{1}{n^{2}+1}\geq \frac{1}{n^2+k}$. So we get that $$\sum_{k=1}^{2n+1} \frac{1}{n^{2}+k} \geq \sum_{k=1}^{2n+1} \frac{1}{n^{2}+2n+1} \ \ \mbox{and} \ \ \sum_{k=1}^{2n+1} \frac{1}{n^{2}+k} \leq \sum_{k=1}^{2n+1}\frac{1}{n^{2}+1}$$ Using this we have $$ \frac{2n^{2}}{n^{2}+2n+1} \leq u_{n} \leq \frac{2n^{2}}{n^{2}+1}$$ Now apply the Squeeze Theorem.

A: How can $u_n$ converge to zero?
$$ u_n = n\sum_{k=1}^{2n+1}\frac{1}{n^2+k}\geq n\cdot\frac{2n+1}{(n+1)^2} \geq 2-\frac{3n+2}{(n+1)^2}.$$
On the other hand,
$$ u_n \leq n\cdot\frac{2n+1}{n^2+1} \leq 2+\frac{n-2}{n^2+1},$$
hence:
$$\lim_{n\to +\infty} u_n = 2.$$
For the second sum, since $\cos x \geq 1-\frac{x^2}{2}$ over $[0,1]$,
$$ v_n \geq \frac{1}{n}\left(n-\frac{1}{2}\sum_{k=0}^{n-1}\frac{1}{n+k}\right)\geq \frac{1}{n}\left(n-\frac{1}{2}\right),$$
but since $\cos x\leq 1-\frac{4x^2}{\pi^2}$ holds over the same interval,
$$ v_n \leq \frac{1}{n}\left(n-\frac{4}{\pi^2}\sum_{k=0}^{n-1}\frac{1}{n+k}\right)\leq \frac{1}{n}\left(n-\frac{4n}{\pi^2(2n-1)}\right),$$
so:
$$\lim_{n\to +\infty} v_n = 1.$$
