Solving a question by mathematical induction

Question

Question : Prove that $$ \sum_{k=1}^n\frac{1}{\sqrt{k}}\le 2\sqrt{n}-1 $$ for all positive integers $n$.

I've been thinking a solution for this question for hours but still can't solve it.

Answer 1

For $n=m$, $$\sum_{k=1}^m\frac1{\sqrt k}\le2\sqrt m-1$$

For $n=m+1,$ $$\sum_{k=1}^{m+1}\frac1{\sqrt k}\le2\sqrt m-1+\frac1{\sqrt{m+1}}$$

It is sufficient to establish $$2\sqrt m-1+\frac1{\sqrt{m+1}}\le2\sqrt{m+1}-1$$

$$\iff\frac1{\sqrt{m+1}}\le2(\sqrt{m+1}-\sqrt m)=\frac2{\sqrt{m+1}+\sqrt m}$$

$$\iff \sqrt{m+1}+\sqrt m\le2\sqrt{m+1}\iff\sqrt m\le\sqrt{m+1}$$ which is true

Answer 2

Base case true. Suppose $n$. Then for $n+1$, it boils down to showing that

$$2\sqrt{n}-1+\frac{1}{\sqrt{n+1}}\leq 2\sqrt{n+1}-1,$$

which is equivalent to

$$\sqrt{n}+\frac{1}{2\sqrt{n+1}}\leq \sqrt{n+1}$$

and upon rearrangement is:

$$\sqrt{n(n+1)}\leq n+1/2$$

and upon squaring both sides (notice that everything is nice and positive), you'll establish the truth of this statement.

Answer 3

Base case should be obvious.

Inductive Step:

$$ \sum_{k = 1}^{n + 1} \dfrac{1}{\sqrt k} = \sum_{k = 1}^{n } \dfrac{1}{\sqrt k} + \dfrac{1}{\sqrt{n + 1}} \le 2\sqrt n - 1 + \dfrac{1}{\sqrt{n + 1}} = \dfrac{ 2\sqrt{ n^2 + n} }{ \sqrt{n + 1} } - 1 \le \dfrac { 2\sqrt{ (n + 1)^2}}{\sqrt{n + 1}} - 1$$

And the Result Follows.