Prove that $$\sum^{n}_{k=1} \frac{1}{\sqrt{k}}>\sqrt{n}$$ for all $n\in \mathbb{N}$ where $n\geq2$. I've already proven the base case for $n=2$, but I don't know how to make the next step.
Is the base case right?
for $n=2$ $$\sum^{2}_{k=1} \frac{1}{\sqrt{k}}=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}=\frac{\sqrt{2}+2}{2}>\sqrt{2}$$
Assume $\sum^{n}_{k=1} \frac{1}{\sqrt{k}}>\sqrt{n}$. Then $$\sum^{n}_{k=1} \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{n+1}}>\sqrt{n} + \frac{1}{\sqrt{n+1}} = \frac{\sqrt{n}\sqrt{n+1}+1}{\sqrt{n+1}}.$$
Now since $n\ge 2$, we have $$\frac{\sqrt{n}\sqrt{n+1}+1}{\sqrt{n+1}} > \frac{\sqrt{n}\sqrt{n}+1}{\sqrt{n+1}} = \frac{n+1}{\sqrt{n+1}} = \sqrt{n+1}.$$
So we have $$\sum^{n+1}_{k=1} \frac{1}{\sqrt{k}} > \sqrt{n+1}.$$
By the AM-GM inequality $$\begin{align*}\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}&\ge n\sqrt[n]{\dfrac{1}{\sqrt{n!}}}>n\sqrt{\dfrac{1}{\sqrt[n]{n^n}}}=n\cdot\dfrac{1}{\sqrt{n}}=\sqrt{n}\end{align*}$$
