Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$

Question

I'm trying to prove $$2(\sqrt{n} - 1) < \sum_{i=1}^n\frac{1}{\sqrt i}$$ (Which is the opposite pretty much of Prove by induction that $\sum_{i = 1}^{n} \frac{1}{\sqrt{i}} \leq 2\sqrt{n} - 1$) And I'm encountering some troubles I fail to see through.

My attempt is by induction, and for $n=1$ it's clear, then if I assume for $k$ I get that for $k+1$: $$2(\sqrt{k+1} - 1) < \sum_{i=1}^{k+1}(\frac{1}{\sqrt i})$$ $$\sum_{i=1}^{k+1}(\frac{1}{\sqrt i}) > 2(\sqrt{k}-1)+\frac{1}{\sqrt{k+1}}$$

Which would mean I would like to find that: $$2(\sqrt{k}-1)+\frac{1}{\sqrt{k+1}} - 2(\sqrt{k+1} - 1) > 0$$ But this is what I get: $$\frac{1}{\sqrt{k+1}}+2\sqrt{k}-2\sqrt{k+1}=\frac{1}{\sqrt{k+1}}+2\frac{\left(\sqrt{k}-\sqrt{k+1}\right)\left(\sqrt{k}+\sqrt{k+1}\right)}{\sqrt{k}+\sqrt{k+1}}=$$ $$ \frac{1}{\sqrt{k+1}}+2\frac{k-k-1}{\sqrt{k}+\sqrt{k+1}}=\frac{1}{\sqrt{k+1}}-\frac{2}{\sqrt{k}+\sqrt{k+1}}=$$ $$ \frac{\sqrt{k}+\sqrt{k+1}-2\sqrt{k+1}}{\sqrt{k}\left(\sqrt{k}+\sqrt{k+1}\right)}<0) $$

So that is not true, how can I find that $\sum_{i=1}^{k+1}(\frac{1}{\sqrt i}) > 2(\sqrt{k+1} - 1)$ if what I get from my induction hypothesis doesn't help me since it's already smaller than what I'm searching for?

Answer 1

You can strengthen your inequality to $$\sum_{i=1}^n{1/\sqrt{i}} > 2(\sqrt{n+1}-1).$$ All you need to do for the induction step is to show $$\frac{1}{\sqrt{k+1}} > 2(\sqrt{k+2}-\sqrt{k+1}).$$ Hint: $$RHS =\frac{2}{\sqrt{k+1}+\sqrt{k+2}}.$$

Answer 2

This is just to illustrate a non-inductive method to proving the inequality:

\begin{align}\sum_{i = 1}^n \frac{1}{\sqrt{i}} &= 2\sum_{i = 1}^n \frac{1}{2\sqrt{i}} \\ & =2\sum_{i = 1}^n \frac{1}{\sqrt{i} + \sqrt{i}} \\ &> 2\sum_{i = 1}^n \frac{1}{\sqrt{i+1} + \sqrt{i}} \\ &= 2\sum_{i = 1}^n (\sqrt{i+1} - \sqrt{i})\\ &= 2[(\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{n+1} - \sqrt{n})]\\ &= 2(\sqrt{n+1} - 1), \end{align}

and $2(\sqrt{n+1} - 1) > 2(\sqrt{n} - 1)$.

Answer 3

Alternative approach: since $f(x)=\sqrt{x}$ is a differentiable and concave function on $\mathbb{R}^+$, $$\delta(x,y) = \left\{\begin{array}{ccc}\frac{f(y)-f(x)}{y-x}&\text{if}&x\neq y\\ f'(x)&\text{if}&x=y\end{array}\right.$$ is a decreasing function with respect to both its parameters. This gives: $$ \frac{1}{2\sqrt{n}}=\delta(n,n)\geq \delta(n,n+1) = \sqrt{n+1}-\sqrt{n}, $$ so, summing both sides of the previous inequality for $n=1,2,\ldots,N$ we get: $$ \sum_{n=1}^{N}\frac{1}{\sqrt{n}}\geq 2\left(\sqrt{N+1}-1\right) $$ as wanted.