<s>[INST] Sum of a series using squeeze theorem How do I find using the Squeeze theorem
$$\lim_{n\to \infty}\sum_{k=1}^n \frac{1}{\sqrt{n^2+k}} \;,$$
using the fact that
$$ \lim_{n\to \infty}\frac{n}{\sqrt{n^2+n}}=1.$$
Thank you very much for your help, 
C.G
 [/INST] Note that
$$\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\leq \frac{n}{\sqrt{n^2+1}}$$
because $\sqrt{n^2+1}\leq\sqrt{n^2+k}$ for $k\geq1$. 
On the other hand, by a similar reasoning,  using that $\sqrt{n^2+n}\geq\sqrt{n^2+k}$ for $k\geq1$:
$$\frac{n}{\sqrt{n^2+n}}\leq\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}$$
so in the end, 
$$\frac{n}{\sqrt{n^2+n}}\leq \sum_{k=1}^n \frac{1}{\sqrt{n^2+k}}\leq \frac{n}{\sqrt{n^2+1}}$$
and use that $$\lim_{n} \frac{n}{\sqrt{n^2+n}}=1=\lim_{n} \frac{n}{\sqrt{n^2+1}}$$
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