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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!

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  • $\begingroup$ If you could share your work for $n=k+1$ case maybe we can direct you to your source of confusion. $\endgroup$ – Sudarsan Sep 23 '13 at 3:49
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(Someone please check for error ... I'm not confident if I got this correct)

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 $c\geq d$.

\begin{align*} \frac{\sqrt{n+1}}{2n}&\geq\frac{\sqrt n-\sqrt{n-1}}2\\ \\ \frac{\sqrt{n+1}}{n}&\geq\sqrt n-\sqrt{n-1}\\ \\ \frac{\sqrt{n+1}}{\sqrt n-\sqrt{n-1}}&\geq n\\ \\ \sqrt{n+1}(\sqrt n+\sqrt{n-1})&\geq n\\ \\ \sqrt{n^2+n}+\sqrt{n^2-1}&\geq n\\\square \end{align*}

(note that $\sqrt{n^2+n}>n$ since $\sqrt{n^2}=n$ for $n\geq1$)

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  • 1
    $\begingroup$ Sounds about right to me.. Thanks! $\endgroup$ – KillerKidz Sep 24 '13 at 14:21
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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 $n\ge2$, $\sqrt{n+1}-1\gt\frac{\sqrt{n}}2$.

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