Prove that inequality $1+\frac{1}{2\sqrt{2}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$ Let $n$ is a natural number. Prove that inequality $$1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$$

My try: $$\frac{1}{n\sqrt{n}}=\frac{\sqrt{n}}{n^2}<\frac{\sqrt{n}}{n\left(n-1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n-1}\right)=\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}\right)\left(\frac{\sqrt{n}+\sqrt{n-1}}{\sqrt{n-1}}\right)<\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}$$
Please check my solution for me and give me some idea. 

Your sum can be written as $\sum_{k=1}^n\frac{1}{k^{3/2}}=1+2^{-3/2}+\sum_{k=3}^n k^{-3/2}$. Since $f(x)=x^{-3/2}$ is decreasing, we can bound the sum from above by an integral, that is,
$$
\sum_{k=3}^n\frac{1}{k^{3/2}}\leq\int_2^n x^{-3/2}\,dx.
$$
Note specifically the lower bound of the integral.
Evaluating the integral on the right yields
$$
\int_2^n x^{-3/2}\,dx=-2x^{-1/2}\bigg|_{x=2}^{x=n}=\frac{2}{\sqrt2}-\frac{2}{\sqrt n}.
$$
Thus, we have
$$
\sum_{k=1}^n\leq1+2^{-3/2}+\frac{2}{\sqrt2}-\frac2{\sqrt n}\leq\frac{2^{3/2}+1+4}{2^{3/2}}.
$$
Clearly $2^{3/2}=2\sqrt2=\sqrt8\leq\sqrt9=3$, so that we have
$$
\frac{2^{3/2}+1+4}{2^{3/2}}\leq\frac{8}{2^{3/2}}=2\sqrt2.
$$
Putting it all together, then, we have
$$
\sum_{k=1}^nk^{-3/2}\leq2\sqrt2.
$$
