Prove $\sum_{1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} \lt 2 $

Question

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.

Answer 1

Because $$\sum_{n=1}^{+\infty}\frac{1}{n(\sqrt{n}+1)}<\sum_{n=1}^{+\infty}\left(\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\right)=\lim_{n\rightarrow+\infty}\left(2-\frac{2}{\sqrt{n+1}}\right)=2$$

Answer 2

$$ \frac{1}{n\sqrt n} = \frac{1}{\sqrt n(n+\sqrt n)}+\frac{1}{n(n+\sqrt n)} $$

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

Answer 3

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 $\sum_{n\geq 1}\frac{1}{n+n\sqrt{n}}<1.7$.