Prove that $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{6}}\cdots>\infty$ .

Question

I need to a fresh solution, but mine

We have that $$\begin{align*} \frac{1}{\sqrt{1}} &+ \frac{1}{\sqrt{2}}+ \frac{1}{\sqrt{3}}+ \frac{1}{\sqrt{4}}+ \frac{1}{\sqrt{5}}+ \frac{1}{\sqrt{6}}+ \frac{1}{\sqrt{7}}\cdots \\ &> \frac{1}{1}+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}\cdots \\ &> \frac{1}{1}+ \frac{1}{2}+ \left ( \frac{1}{4}+ \frac{1}{4} \right )+ \left ( \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8} \right )\cdots \\ &= 1+ \frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}\cdots \\ &= \infty \end{align*}$$

Answer 1

Because $$\frac{1}{\sqrt{n}}>\frac{2}{\sqrt{n}+\sqrt{n+1}}=2(\sqrt{n+1}-\sqrt{n})$$ and the telescopic summation.

Answer 2

An alternative argument, appealing also to the fact that the harmonic series diverges:

$$\underbrace{\frac{1}{\sqrt 1}}_{\displaystyle \ge 1} + \underbrace{\frac{1}{\sqrt 2} + \cdots +\frac{1}{\sqrt 4}}_{\displaystyle \ge \frac{1}{\sqrt 4} = \frac 1 2} + \underbrace{\frac{1}{\sqrt 5} + \cdots +\frac{1}{\sqrt 9}}_{\displaystyle \ge \frac{1}{\sqrt 9} = \frac 1 3} + \cdots$$

In general: let $k^2,(k+1)^2$ be consecutive perfect squares. Then, since $\sqrt x \ge 0$, you'll have

$$\frac{1}{\sqrt{k^2+1}} + \cdots +\frac{1}{\sqrt{(k+1)^2}} \ge \frac{1}{\sqrt{(k+1)^2}}=\frac 1 {k+1}$$

Doing this over and over again throughout the sum allows one to show that

$$\sum_{k=1}^\infty \frac{1}{\sqrt k} \ge \sum_{k=1}^\infty \frac 1 k$$

but the latter diverges.