Question

Prove that for every positive integer n, $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots +\frac{1}{\sqrt{n}}>2(\sqrt{n+1}-1)$

Short answer

n is a positive integer, then $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots +\frac{1}{\sqrt{n}}>2(\sqrt{n+1}-1)$

Step-by-step solution

Step: 1

If n=1,

$\begin{array}{r}1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots +\frac{1}{\sqrt{1}}>2(\sqrt{1+1}-1)\\ 2\sqrt{2}-2\approx 0.8284\end{array}$

it is true for n=1.

Step: 2

Let P(k) be true.

$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots +\frac{1}{\sqrt{k}}>2(\sqrt{k+1}-1)$

We need to prove that P(k) is true.

Step: 3

$\begin{array}{r}1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots +\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}\\ \frac{1}{\sqrt{k+1}}\ge 2(\sqrt{k+2}-1)-2(\sqrt{k+1}-1)\\ =2\sqrt{k+2}-2-2\sqrt{k+1}+2\\ =2\sqrt{k+2}-2\sqrt{k+1}\\ =2(\sqrt{k+2}-\sqrt{k+1})\\ >2(\sqrt{k+1}-1)\end{array}$

It is true for P(k) is true