How can I solve the following sequence [closed]
Hint
Prove that $$\frac{n}{\sqrt{n^2+n}}\leq \sum\frac{1}{\sqrt{n^2+k}}\leq\frac{n}{\sqrt{n^2}}$$
$$\frac{n}{\sqrt{n^2+n}}\leq y_n\leq\frac{n}{\sqrt{n^2}}$$
Then use squeeze lemma
It is a increasing sequence
Knowing that $$\frac{n}{\sqrt{n^2+n}}\leq y_n\leq\frac{1}{n}$$
We have that $y_n\leq\frac{1}{n}$ and $\frac{n+1}{\sqrt{(n+1)^2+n+1}}\leq y_{n+1}$
And easy you can show that $$\frac{1}{n}\leq \frac{n+1}{\sqrt{(n+1)^2+n+1}}$$
So $$y_
n\leq y_{n+1}$$