Proof by induction with square root in denominator: $\frac1{2\sqrt1}+\frac1{3\sqrt2}+\dots+ \frac1{(n+1)\sqrt n} < 2-\frac2{\sqrt{(n+1)}}$ I need to prove $\frac{1}{2\sqrt1} + \frac{1}{3\sqrt2} + ... + \frac{1}{(n+1)\sqrt n} < 2 - \frac{2}{\sqrt{(n+1)}}$
by induction for every $n \in  \mathbb{N} $. 
Please help, I am stuck on this from the past 2 days..
Thanks.
 A: Without induction you can show it immediately as follows:
$$\sum_{k=1}^n\frac{1}{\sqrt{k}(k + 1)}\leq \sum_{k=1}^n\frac{1}{k \sqrt{k}} < \int_1^{n+1}\frac{1}{x^{\frac{3}{2}}}\,dx =\left[-2\frac{1}{\sqrt{x}} \right]_1^{n+1} = 2 -\frac{2}{\sqrt{n+1}}$$
A: Great answer by trancelocation, but in case you still want it, here is how to do induction step for an inductive proof.  First we note the following general rule of quadratics:
$$(n + \tfrac{a+b}{2})^2 - (n+a)(n+b) = (\tfrac{a-b}{2})^2.$$
Using this rule with $a=1$ and $b=2$, we have:
$$(n+\tfrac{3}{2})^2 > (n+1)(n+2).$$
Rearranging this inequality gives:
$$\frac{(n+\tfrac{3}{2})^2}{n+1} > n+2.$$
Our inductive step can now be accomplished as follows:
$$\begin{equation} \begin{aligned}
\sum_{k=1}^{n+1} \frac{1}{(k+1)\sqrt{k}} 
&= \sum_{k=1}^{n} \frac{1}{(k+1)\sqrt{k}} + \frac{1}{(n+2)\sqrt{n+1}} \\[6pt]
&< 2 - \frac{2}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{n+1}} \\[6pt]
&= 2 - \frac{2(n+2)-1}{(n+2)\sqrt{n+1}} \\[6pt]
&= 2 - \frac{2n+3}{(n+2)\sqrt{n+1}} \\[6pt]
&= 2 - \frac{2}{(n+2)} \cdot \frac{n+\tfrac{3}{2}}{\sqrt{n+1}} \\[6pt]
&< 2 - \frac{2}{(n+2)} \cdot \sqrt{n+2} \\[6pt]
&= 2 - \frac{2}{\sqrt{n+2}} . \\[6pt]
\end{aligned} \end{equation}$$
