<s>[INST] Proving $\sum_{n=1}^{+\infty}\frac{1}{n\sqrt{n+2}}<2$ without calculus I would like to prove the following inequality without using calculus :
$$
\sum_{n=1}^{+\infty}\frac{1}{n\sqrt{n+2}}<2
$$
Any hint?
Thank you very much!
 [/INST] We have 
\begin{align*}
\frac{1}{n\sqrt{n+2}}\leq\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n+1}},\qquad n\geq 2,
\end{align*}
and here is why: First note that $\sqrt{n^2-1}\leq n$ for all $n\geq 1$. Moreover, $\sqrt{n+1}-\sqrt{n-1}=\frac{2}{\sqrt{n+1}+\sqrt{n-1}}\geq \frac{1}{\sqrt{n+2}}$. From this we get
\begin{align*}
\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n-1}}{\sqrt{n^2-1}}\geq\frac{1}{n\sqrt{n+2}},
\end{align*}
as claimed. 
Now, we obtain
\begin{align*}
\sum_{n=1}^\infty\frac{1}{n\sqrt{n+2}}&\leq\sum_{n=1}^6\frac{1}{n\sqrt{n+2}}+\sum_{n=7}^\infty\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n+1}}\right) \\ &=\frac{1}{4}+\frac{1}{\sqrt{3}}+\frac{1}{3 \sqrt{5}}+\frac{5}{4 \sqrt{6}}+\frac{6}{5 \sqrt{7}}+\frac{1}{12 \sqrt{2}}<2.
\end{align*}
</s>