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!
 A: 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*}
A: Hint $n\sqrt{n+2}>n^{3/2}$, compare with $\int_1^{\infty}x^{-3/2}dx$.
A: Disclaimer: this is a pretty technical approach. The inverse Laplace transform of $\frac{1}{x\sqrt{x+2}}$ is given by $\frac{1}{\sqrt{2}}\text{Erf}(\sqrt{2s})$, hence the have the identity
$$ \sum_{n\geq 1}\frac{1}{n\sqrt{n+2}} = \frac{1}{\sqrt{2}}\int_{0}^{+\infty}\frac{\text{Erf}(\sqrt{2s})}{e^s-1}\,ds\stackrel{s\mapsto u^2/2}{=}\frac{1}{\sqrt{2}}\int_{0}^{+\infty}\frac{u\,\text{Erf}(u)}{e^{u^2/2}-1}\,du $$
and the wanted inequality can be recovered from the continued fraction expansion of the error function and the Cauchy-Schwarz inequality.

A simpler approach is to notice that $\frac{1}{n\sqrt{n+2}}< \frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}$ and $\frac{1}{n\sqrt{n+2}}> \frac{2}{\sqrt{n+\frac{1}{6}}}-\frac{2}{\sqrt{n+\frac{7}{6}}}$ hold for any $n\geq 1$, hence by creative telescoping
$$ \sum_{n\geq 1}\frac{1}{n\sqrt{n+2}}\in \frac{1}{\sqrt{3}}+\left(\sqrt{\frac{24}{13}},\sqrt{2}\right)=(1.93608271\ldots,1.99156383\ldots). $$
