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.
 A: The following statements are equivalent. The first statement shows that $P(n)<2\sqrt{n}\Rightarrow P(n+1)<2\sqrt{n+1}$ and the last statement is evidently true:
$2\sqrt{n}+\frac{1}{\sqrt{n+1}}\leq2\sqrt{n+1}$
$2\sqrt{n\left(n+1\right)}+1\leq2\left(n+1\right)$
$2\sqrt{n\left(n+1\right)}\leq2n+1$
$4n\left(n+1\right)\leq4n^{2}+4n+1$
A: Add $\frac{1}{\sqrt{k+1}} $ to both sides of $P(k)$ and then show
$$2\sqrt{k}+\frac{1}{\sqrt{k+1}} < 2\sqrt{k+1}$$
A: First note that $$\frac{1}{\sqrt{n}} = \frac{2}{\sqrt{n}+\sqrt{n} }\leq \frac{2}{\sqrt{n}+\sqrt{n-1}} \overset{(1)}{=} 2\left(\sqrt{n}-\sqrt{n-1} \right)$$
$(1)$ follows by multiplication with conjugate.
Now sum to get: $$\sum_ {k=1}^{n} \frac{1}{\sqrt{k}} \leq \sqrt{n}-\sqrt{0}= \sqrt{n}$$
A: $P(k+1)=P(k)+\dfrac{1}{\sqrt{k+1}}\leq 2\sqrt{k}+\dfrac{1}{\sqrt{k+1}}=\dfrac{2\sqrt{k}\sqrt{k+1}+1}{\sqrt{k+1}}$, 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 $P(k+1)\leq 2\sqrt{k+1}$.  
A: If you want to take a look on the Riemann's sum method : 
$\forall n > 1$, we have $$\int_{n-1}^n \frac{dt}{\sqrt{t}} \ge (n-(n-1))\cdot \underset{x \in [n-1,n]}{\min} \frac{1}{\sqrt{x}} = \frac{1}{\sqrt{n}}  $$
Hence, $$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} \le 1+ \int_{1}^2 \frac{dt}{\sqrt{t}} + \int_{2}^3 \frac{dt}{\sqrt{t}} + ... + \int_{n-1}^n \frac{dt}{\sqrt{t}} = 1+\int_{1}^n\frac{dt}{\sqrt{t}}  $$
$$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}}  \le 1+\left[2\sqrt{t}\right]_1^n  = 1+ (2\sqrt{n} -2 )= 2\sqrt{n}  -1 <  2\sqrt{n} $$ 
The difference can be seen by comparing the Right Riemann Sum of the function $t \rightarrow \frac{1}{\sqrt{t}} $ to its integral on $[1,n]$.
A: 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.
