Proving $\sum_{k=1}^n\frac1{\sqrt k}<2\sqrt n$ by induction [closed]

I am just starting out learning mathematical induction and I got this homework question to prove with induction but I am not managing. $$\sum_{k=1}^n\frac1{\sqrt k}<2\sqrt n$$ Please help!

Solution 1:

One may observe that, for $n>1$, $$ \begin{align} \sum_{k=1}^n \frac{1}{2\sqrt{k}}&= \sum_{k=1}^n \frac{1}{\sqrt{k} + \sqrt{k}} \\&\color{red}{<} \sum_{k=1}^n \frac{1}{\sqrt{k} + \sqrt{k-1}} \\&=\sum_{k=1}^n \frac{\sqrt{k} - \sqrt{k-1}}{(\sqrt{k} + \sqrt{k-1})(\sqrt{k} - \sqrt{k-1})} \\&= \sum_{k=1}^n (\sqrt{k} - \sqrt{k-1}) \\& = \sqrt{n} \end{align} $$ giving that

$$ \sum_{k=1}^n \frac{1}{\sqrt{k}}<2 \sqrt{n}, \quad n=1,2,3,\cdots. $$

Solution 2:

Induction step

Suppose for some $n\ge1$, one has $\;\displaystyle\sum_{i=1}^n\frac1{\sqrt k}<2\sqrt n$. Then $$\sum_{i=1}^{n+1}\frac1{\sqrt k}=\sum_{i=1}^{n}\frac1{\sqrt k}+\frac1{\sqrt{n+1}}<2\sqrt n+\frac1{\sqrt{n+1}}. $$ Now \begin{align} 2\sqrt n+\frac1{\sqrt{n+1}}<2\sqrt{n+1}&\iff\frac1{2\sqrt{n+1}} <\sqrt{n+1}-\sqrt n=\frac 1{\sqrt{n+1}+\sqrt n}\\ &\iff 2\sqrt{n+1}>\sqrt{n+1}+\sqrt n\iff\sqrt{n+1}>\sqrt n. \end{align}

Without induction:

$\displaystyle\sum_{i=1}^n\frac1{\sqrt k}$ is a lower Riemann sum for the (improper) Riemann integral $$\displaystyle\int_0^n\frac{\mathrm d\mkern1mu x}{\sqrt x}=2\sqrt x\,\biggr\rvert_0^n=2\sqrt n.$$