<s>[INST] Prove $\sum_{1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} \lt 2 $ 
Prove that $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} \lt 2$$ 

I have found that $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} < \pi / 2 $ with integrating from $1$ to infinity but integral isn’t allowed.
 [/INST] Let us go with creative telescoping. As $n\to +\infty$,
$$ \frac{1}{n+n\sqrt{n}}=\frac{1}{n\sqrt{n}}-\frac{1}{n^2}+\frac{1}{n^2\sqrt{n}}+O\left(\frac{1}{n^3}\right) $$
while
$$ \frac{2}{\sqrt{n-\frac{7}{6}}}-\frac{2}{\sqrt{n-\frac{1}{6}}}=\frac{1}{n\sqrt{n}}+\frac{1}{n^2\sqrt{n}}+\frac{95}{96 n^2\sqrt{n}}+o\left(\frac{1}{n^3}\right) $$
$$ \frac{1}{n-1}-\frac{1}{n} = \frac{1}{n^2}+\frac{1}{n^3}+o\left(\frac{1}{n^3}\right) $$
hence by setting $f(n)=\frac{2}{\sqrt{n-\frac{7}{6}}}-\frac{1}{n-1}$ we have that $f(n)-f(n+1)$ is an excellent approximation of $\frac{1}{n+n\sqrt{n}}$ for any $n\geq 4$, such that
$$ \sum_{n\geq 1}\frac{1}{n+n\sqrt{n}}\approx \sum_{n=1}^{3}\frac{1}{n+n\sqrt{n}}+f(4) =1.6839589\ldots$$
is accurate up to the second figure. In particular $\sum_{n\geq 1}\frac{1}{n+n\sqrt{n}}<1.7$.
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