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.

$$
\frac{1}{n\sqrt n} = \frac{1}{\sqrt n(n+\sqrt n)}+\frac{1}{n(n+\sqrt n)}
$$
hence
$$
 \frac{1}{\sqrt n(n+\sqrt n)}<\frac{1}{n^{\frac 32}}
$$
NOTE
$$
\sum_{n=3}^{\infty} \frac{1}{n^{\frac 32}} < 1.24
$$
now adding $\frac{1}{\sqrt 1(1+\sqrt 1)}+\frac{1}{\sqrt 2(2+\sqrt 2)}+\sum_{n=3}^{\infty} \frac{1}{n^{\frac 32}} < 1.95$
