<s>[INST] 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.
 [/INST] $P(k+1)=P(k)+\dfrac{1}{\sqrt{k+1}}\leq 2\sqrt{k}+\dfrac{1}{\sqrt{k+1}}=\dfrac{2\sqrt{k}\sqrt{k+1}+1}{\sqrt{k+1}}$, in the other hand we have $2\sqrt{k}\sqrt{k+1}\leq k+k+1=2k+1$  (using $2ab\leq a^2+b^2$), now we get $2\sqrt{k}\sqrt{k+1}+1\leq 2(k+1)$, and finally $\dfrac{2\sqrt{k}\sqrt{k+1}+1}{\sqrt{k+1}}\leq 2\sqrt{k+1}$, so $P(k+1)\leq 2\sqrt{k+1}$.  
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