# Thread: need help with Induction

1. ## need help with Induction

Sum of n=1 to N of 1/sqrt(n) > sqrt(N) for N>= 2.

Thanks for any help.

2. Originally Posted by alpha
Show using an inductive or non-inductive argument that $\displaystyle \sum_{n=1}^N\frac{1}{\sqrt{n}} > \sqrt{N}$ for $\displaystyle N\geq 2$.
Thanks for any help.
Clearly $\displaystyle N<N+1$ for all $\displaystyle N\geq 2$. Then

$\displaystyle \sqrt{N}<\sqrt{N+1}$

$\displaystyle \implies$ $\displaystyle N<\sqrt{N}\sqrt{N+1}$

$\displaystyle \implies$ $\displaystyle N+1-\sqrt{N}\sqrt{N+1}<1$

$\displaystyle \implies$ $\displaystyle \sqrt{N+1}-\sqrt{N}<\frac{1}{\sqrt{N+1}}$.

So if $\displaystyle \sqrt{N}<\sum_{n=1}^N\frac{1}{\sqrt{n}}$, then

$\displaystyle \sqrt{N}+\sqrt{N+1}-\sqrt{N}<\sum_{n=1}^N\frac{1}{\sqrt{n}}+\frac{1}{\ sqrt{N+1}}$

$\displaystyle \implies$ $\displaystyle \sqrt{N+1}<\sum_{n=1}^{N+1}\frac{1}{\sqrt{n}}$.

Since furthermore $\displaystyle \sqrt{2}<1+\frac{1}{\sqrt{2}}=\sum_{n=1}^2\frac{1} {\sqrt{n}}$, then $\displaystyle \sum_{n=1}^N\frac{1}{\sqrt{n}} > \sqrt{N}$ for all $\displaystyle N\geq 2$.