Inequality involving a finite sum this is my first post here so pardon me if I make any mistakes.
I am required to prove the following, through mathematical induction or otherwise:
$$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} < 2{\sqrt{n}}$$
I tried using mathematical induction through:
$Let$ $P(n) = \frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} < 2{\sqrt{n}}$
$Since$ $P(1) = \frac{1}{\sqrt1} < 2{\sqrt{1}}, and$ $P(k) = \frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{k}} < 2{\sqrt{k}},$
$P(k+1) = \frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{k}}+ \frac{1}{\sqrt{k+1}} < 2{\sqrt{k+1}}$
Unfortunately, as I am quite new to induction, I couldn't really proceed from there. Additionally, I'm not sure how to express ${\sqrt{k+1}}$ in terms of ${\sqrt{k}}$ which would have helped me solve this question much more easily. I am also aware that this can be solved with Riemann's Sum (or at least I have seen it being solved in that way) but I do not remember nor quite understand it.

We have a stronger inequality.  Note that
$$\sqrt{x-\frac{1}{2}}+\sqrt{x+\frac{1}{2}}<2\sqrt{x}$$
for all $x\geq \dfrac{1}{2}$.  Thus, for each $x>\dfrac12$, we get
$$\frac{1}{\sqrt{x}}< \frac{2}{\sqrt{x-\frac12}+\sqrt{x+\frac{1}{2}}}=2\,\left(\sqrt{x+\frac12}-\sqrt{x-\frac12}\right)\,.$$
That is, for every nonnegative integer $n$,
$$\sum_{k=1}^n\,\frac{1}{\sqrt{k}}\leq 2\,\sum_{k=1}^n\,\left(\sqrt{k+\frac12}-\sqrt{k-\frac12}\right)=2\,\left(\sqrt{n+\frac12}-\sqrt{\frac12}\right)\,.$$
The equality holds if and only if $n=0$.  This is a sharper inequality than the required inequality as
$$\sqrt{x+y}\leq \sqrt{x}+\sqrt{y}$$
for any $x,y\geq 0$ (with equality iff $x=0$ or $y=0$).
On the other hand, you can show that
$$2\sqrt{x}\leq \sqrt{x-\frac{7}{16}}+\sqrt{x+\frac{9}{16}}$$
for every $x\geq 1$ (with equality case $x=1$).  This gives
$$\sum_{k=1}^n\,\frac{1}{\sqrt{k}}\geq 2\,\left(\sqrt{n+\frac{9}{16}}-\frac{3}{4}\right)$$
for all integers $n\geq 0$ with equality cases $n=0$ and $n=1$.  That is,
$$2\,(\sqrt{n+1}-1)\leq 2\,\left(\sqrt{n+\frac{9}{16}}-\frac{3}{4}\right)\leq \sum_{k=1}^n\,\frac{1}{\sqrt{k}}\leq 2\,\left(\sqrt{n+\frac12}-\sqrt{\frac12}\right)\leq 2\sqrt{n}$$
for every nonnegative integer $n$.  If $n=0$, then every inequality is an equality.  If $n=1$, then only the second inequality from the left becomes an inequality.  For $n>1$, all inequalities are strict.
