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. 
 A: 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
$$
\sum_{k=1}^nk^{-3/2}\leq2\sqrt2.
$$
A: I think your way gets something wrong.
By your work:
For all $n\geq2$ we obtain:
$$\frac{1}{n\sqrt{n}}<\left(\frac{1}{\sqrt{n-1}}-\frac{1}
{\sqrt{n}}\right)\frac{\sqrt{n}+\sqrt{n-1}}{\sqrt{n-1}}=$$
$$=\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\left(1+\sqrt{\frac{n}{n-1}}\right)\leq\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)(1+\sqrt2).$$
Thus, $$\sum_{k=1}^n\frac{1}{k\sqrt{k}}=1+\sum_{k=2}^n\frac{1}{k\sqrt{k}}\leq1+(1+\sqrt2)\left(1-\frac{1}{\sqrt{n}}\right).$$
Id est, it's enough to prove that
$$1+(1+\sqrt2)\left(1-\frac{1}{\sqrt{n}}\right)<2\sqrt2,$$ which is wrong for $n\rightarrow+\infty$.
A: We may consider that $f(x)=\frac{1}{\sqrt{x}}$ is a convex function on $\mathbb{R}^+$, hence for any $n\in\mathbb{N}^+$ we have
$$ f(n-1/2)-f(n+1/2) \geq -f'(n) $$
or
$$ \frac{1}{\sqrt{n-1/2}}-\frac{1}{\sqrt{n+1/2}} \geq \frac{1}{2n\sqrt{n}} $$
such that by creative telescoping
$$ \sum_{n=1}^{N}\frac{1}{n\sqrt{n}} \leq 2\left(\frac{1}{\sqrt{1-1/2}}-\frac{1}{\sqrt{N+1/2}}\right) < 2\sqrt{2}.$$
The inequality is pretty sharp since $\zeta\left(\frac{3}{2}\right)=2.612375\ldots$
The same approach proves $\zeta\left(\frac{3}{2}\right)\leq 1+2\sqrt{\frac{2}{3}}<2.633$.
A: The original inequality cannot be proved directly by induction but by the solution given by Jack D'Aurizio is interesting to note that we can also proceed by induction proving the stronger result
$$1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}-\frac{2}{\sqrt{n+\frac12}}<2\sqrt{2}$$
indeed it is true for $n=1$ since
$$1<2\sqrt 2-2\sqrt{\frac23}\approx 1.20$$
and by induction step we obtain
$$1+\frac{1}{2\sqrt{2}}+...+\frac{1}{(n+1)\sqrt{n+1}}\stackrel{Hyp.}<2\sqrt{2}-\frac{2}{\sqrt{n+\frac12}}+\frac{1}{(n+1)\sqrt{n+1}}\stackrel{?}<2\sqrt{2}-\frac{2}{\sqrt{n+\frac32}}$$
which is true indeed
$$\frac{1}{(n+1)\sqrt{n+1}}<\frac{2}{\sqrt{n+\frac12}}-\frac{2}{\sqrt{n+\frac32}} \iff \frac{1}{2n\sqrt{n}}<\frac{1}{\sqrt{n-\frac12}}-\frac{1}{\sqrt{n+\frac12}}$$
and by convexity of $f(x)=\frac1{\sqrt {n-x}}-\frac1{\sqrt {n+x}}$ we have
$$f'(x)=\frac1{\sqrt {(n-x)^3}}+\frac1{\sqrt {(n+x)^3}}$$
$$\frac1{2n\sqrt {n}}<f'(0)=\frac2{n\sqrt {n}}\le f(1/2)-f(0)=\frac{1}{\sqrt{n-\frac12}}-\frac{1}{\sqrt{n+\frac12}}$$
