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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:

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

Community
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asd
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  • Have you tried integration? – TravisJ Feb 19 '15 at 15:20
  • @TravisJ I don't know how to integrate (I haven't reached that part of the book yet (: ). – asd Feb 19 '15 at 15:24
  • Have you tried $n=1$: $\frac{1}{\sqrt{1}}=1\not<2\cdot\sqrt{1}-1$ – Pim Feb 19 '15 at 15:25
  • @Pim Sorry. I had forgotten to include the constraints for $n$. – asd Feb 19 '15 at 15:28
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    You mentioned you tried telescoping - showing your work would have helped. Sometimes all it takes is to consider keeping one or two terms outside the telescoping part to get much better accuracy. – Macavity Feb 19 '15 at 15:38
  • @Macavity Now I've put my attempt. I tried to start the summation with $k=2$ to end up with a $\sqrt{2}-1$ on the RHS, but that changed the LHS too. – asd Feb 19 '15 at 16:00
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    @Jazz starting from $i=2$ is the only difference between your attempt and Daniels excellent answer below! – Macavity Feb 19 '15 at 16:14

3 Answers3

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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

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

Daniel Fischer
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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 $n\geq2$ $$\sum_{k=1}^{n}\frac{1}{\sqrt{k}}<\sqrt{n}+\int_{1}^{n}\frac{t}{2t^{3/2}}dt=2\sqrt{n}-1.$$

Marco Cantarini
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  • Thanks. Unfortunately I do not understand that part with the integral |:, but I hope to do so soon (:. – asd Feb 19 '15 at 15:51
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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: enter image description here

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.

Kim Jong Un
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