# Prove $\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{4} + \frac{\sqrt{4}}{6} + \cdots + \frac{\sqrt{n+1}}{2n} > \frac{\sqrt{n}}{2}$ by induction

Prove by induction that for all $n > 0$,

$$\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{4} + \frac{\sqrt{4}}{6} + \cdots + \frac{\sqrt{n+1}}{2n} > \frac{\sqrt{n}}{2}$$

I have done the basis step, where $n = 1$ and showed that L.H.S is > R.H.S

For inductive step, I assume $n = k$ & L.H.S > R.H.S is true.

However, I am stuck at showing how for $k + 1$, it is also true for L.H.S > R.H.S

Any help will be much appreciated. Thanks!

• If you could share your work for $n=k+1$ case maybe we can direct you to your source of confusion. – Sudarsan Sep 23 '13 at 3:49

Since for the base case $n=1$: $\dfrac{\sqrt2}2>\dfrac{\sqrt 1}2$, you just have to prove that $\dfrac{\sqrt{n+1}}{2n}\geq\dfrac{\sqrt{n}}2-\dfrac{\sqrt{n-1}}2$ for $n>1$ since, if $a>b$ then $a+c>b+d$ for $c\geq d$.
(note that $\sqrt{n^2+n}>n$ since $\sqrt{n^2}=n$ for $n\geq1$)
By cross-multiplying, it can easily be seen that \begin{align} \frac{\sqrt{k+1}}{2k} &\gt\frac1{\sqrt{k+1}+\sqrt{k\,}}\\[6pt] &=\sqrt{k+1}-\sqrt{k\,} \end{align} Summing gives $$\sum_{k=1}^n\frac{\sqrt{k+1}}{2k} \gt\sqrt{n+1}-1$$ This is not the inequality given, but for $n\ge2$, $\sqrt{n+1}-1\gt\frac{\sqrt{n}}2$.