Improving an Inequality

Problem

In an old Russian collection of problems I found an elementary inequality that was easy to prove but gave a nice example of a telescoping summation. I also add a proof by induction. The proofs make it clear that the inequality is rather weak which led to the question of improving it, i.e., making it stronger, less lax. Here's what transpired.

Prove that, for integer $n\ge 1,\;$

$\displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}\gt 2(\sqrt{n+1}-1).$

Solution 1

First proof is by induction. Denote the sum on the left as $S_n.\;$ Then $S_1=1\gt 2{\sqrt{2}-1}.\;$ For the inductive step, assume $S_n\gt 2(\sqrt{n+1}-1).\;$ We wish to show that $S_{n+1}\gt 2(\sqrt{n+2}-1).\;$

Observe that $\displaystyle S_{n+1}=S_n+\frac{1}{n+1}.\;$ Thus, suffice it to show that $\displaystyle \frac{1}{\sqrt{n+1}}\gt 2(\sqrt{n+2}-\sqrt{n+1}).\;$ This is indeed so:

$\displaystyle\begin{align} 2(\sqrt{n+2}-\sqrt{n+1}) &= \frac{2}{\sqrt{n+2}+\sqrt{n+1}}\\ &\lt \frac{2}{\sqrt{n+1}+\sqrt{n+1}}\\ &= \frac{1}{\sqrt{n+1}}. \end{align}$

Solution 2

As we just verified, for any $k\ge 1,\;$ $\displaystyle \frac{1}{\sqrt{k}}\gt 2(\sqrt{k+1}-\sqrt{k}).\;$ Let's write all these inequality in a column:

$\displaystyle\begin{align} \frac{1}{\sqrt{1}}&\gt 2(\sqrt{1+1}-\sqrt{1})=2(\sqrt{2}-1)\\ \frac{1}{\sqrt{2}}&\gt 2(\sqrt{2+1}-\sqrt{2})\\ \frac{1}{\sqrt{3}}&\gt 2(\sqrt{3+1}-\sqrt{3})\\ &\cdots\\ \frac{1}{\sqrt{n}}&\gt 2(\sqrt{n+1}-\sqrt{n}).\\ \end{align}$

Summing up gives exactly the required inequality. It is obvious that the telescoping is simply performs several induction steps at once.

Improvement

The inequality we just proved is not very strong, indeed, say, $S_{100}\approx 18.5896,\;$ whereas $2(\sqrt{101}-1)\approx 18.09975\;$ - difference of about $0.5\;$ and, clearly, the gap only grows with every increase in $n.\;$ So we may pose a question of strengthening the inequality. Observe that much of the gap comes from the very first step: $1\gt 2(\sqrt{2}-1)\approx 0.82843.\;$ So this is where we may start:

What is the largest $\alpha\;$ for which $1\gt 2(\sqrt{1+\alpha}-1)?$

Rather obviously, since $y=\sqrt{x}\;$ is strictly increasing we'd like to find $\alpha\;$ for which $2(\sqrt{1+\alpha}-1)=1\;$ while for lesser $\alpha\text{'s}\;$ we wold have an inequality. The sought $\displaystyle\alpha =\frac{5}{4}.\;$ Thus we may want to consider the problem

Prove that, for integer $n\ge 2,\;$

$\displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}\gt \sqrt{4n+5}-2.$

The telescoping should still work:

$\displaystyle\begin{align} \frac{1}{\sqrt{1}}&\ge \sqrt{4+5}-2\\ \frac{1}{\sqrt{2}}&\gt \sqrt{8+5}-\sqrt{9}\\ \frac{1}{\sqrt{3}}&\gt \sqrt{12+5}-\sqrt{13}\\ &\cdots\\ \frac{1}{\sqrt{n}}&\gt \sqrt{4n+9}-\sqrt{4n+5}.\\ \end{align}$

It does work because, for $\sqrt{k}\gt 1,\;$ $\displaystyle\frac{1}{k}\gt\sqrt{4k+9}-\sqrt{4k+5}.\;$ Indeed, as before,

$\displaystyle\begin{align} \sqrt{4k+9}-\sqrt{4k+5} &= \frac{4}{\sqrt{4k+9}+\sqrt{4k+5}}\\ &\lt \frac{4}{\sqrt{4k+4}+\sqrt{4k+4}}\\ &= \frac{1}{\sqrt{k+1}}. \end{align}$

We can check: $(\sqrt{4n+5}-2)(100)\approx 18.1246.\;$ Well, some improvement.

Now, $\displaystyle\int_{1}^{n+1}\frac{dx}{\sqrt{x}}=2(\sqrt{n+1}-1).\;$ This is probably where the original right-hand side came from. On the other hand, the upper Riemann sums give $\displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}.\;$ This is where the original left-hand side came from. (This could be considered as the third proof of the inequality.) Using this information it may be possible to improve the inequality further. May you try?

 

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

 62816537

Search by google: