Can the inequality $\sum\limits_{i=1}^n\frac{1}{\sqrt{i}} < 2\sqrt{n} - 1$ be proved without induction?

Question

Maybe some of you have seen one of the posts where the inequality $\sum\limits_{i=1}^n\frac{1}{\sqrt{i}} < 2\sqrt{n}$ is proved by induction (here and here). It can be proved without induction too, but the one with which I'm struggling to prove directly is this:

$$\sum_{i=1}^n\frac{1}{\sqrt{i}} < 2\sqrt{n} - 1 \qquad \text{For }n \in \mathbb{N}, n\geq2.$$

By applying the telescoping property for sums I reached the same inequality with just the $2\sqrt{n}$, but I don't know how to deal with the $1$ there.

My attempt was this:

$$ \begin{align*} \frac{1}{2\sqrt{i}} = \frac{1}{\sqrt{i}+ \sqrt{i}} &< \frac{1}{\sqrt{i}+ \sqrt{i-1}} = \sqrt{i} -\sqrt{i-1}\\ \sum_{i=1}^n\frac{1}{2\sqrt{i}} &< \sum_{i=1}^n[\sqrt{i} -\sqrt{i-1}]\\ \sum_{i=1}^n\frac{1}{2\sqrt{i}} &< \sqrt{n} -\sqrt{1-1}\\ \sum_{i=1}^n\frac{1}{\sqrt{i}} &< 2\sqrt{n} \end{align*} $$

Answer 1

With telescoping, write

$$\sum_{i=1}^n \frac{1}{\sqrt{i}} = 1 + \sum_{i=2}^n \frac{1}{\sqrt{i}}.$$

Then use

$$2(\sqrt{i} - \sqrt{i-1}) = \frac{2}{\sqrt{i} + \sqrt{i-1}} > \frac{1}{\sqrt{i}}$$

to conclude

$$\sum_{i=1}^n \frac{1}{\sqrt{i}} < 1 + 2\sum_{i=2}^n (\sqrt{i} - \sqrt{i-1}) = 1 + 2(\sqrt{n}-1) = 2\sqrt{n} - 1.$$

Answer 2

Using Abel summation formula http://en.wikipedia.org/wiki/Abel%27s_summation_formula you have $$\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=\frac{n}{\sqrt{n}}+\int_{1}^{n}\frac{\left\lfloor t\right\rfloor }{2t^{3/2}}dt$$ where $\left\lfloor t\right\rfloor$ is the integer part of $t.$ So, if $n\geq2$ $$\sum_{k=1}^{n}\frac{1}{\sqrt{k}}<\sqrt{n}+\int_{1}^{n}\frac{t}{2t^{3/2}}dt=2\sqrt{n}-1.$$

Answer 3

Please take this as a comment. By considering the convexity of $i\mapsto\frac{1}{\sqrt{i}}$, you may use Jensen's inequality to find a lower bound for your sum: $$ \sum_i\frac{1}{\sqrt{i}}=n\sum_i\frac{1}{n}\frac{1}{\sqrt{i}}\geq n\frac{1}{\sqrt{\sum_ii/n}}=\frac{n}{\sqrt{(n+1)/2}}\cdot $$ This lower bound and the upper bound provided in the question look like this:

Because $$ \frac{n}{\sqrt{(n+1)/2}}=\sqrt{2(n+1)}-\frac{1}{\sqrt{(n+1)/2}}, $$ you can use this lower bound to infer the divergence of the series.