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I would like to prove the following inequality without using calculus : $$ \sum_{n=1}^{+\infty}\frac{1}{n\sqrt{n+2}}<2 $$

Any hint?

Thank you very much!

Vic. Utap
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  • @XanderHenderson Since 1/(n*sqrt(n+2)) < 1/(n^{3/2}), the value of my infinite sum is bounded by the value of the Riemann zeta function in 1.5, but this value is higher than 2 ... – Vic. Utap Mar 14 '18 at 17:04
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    The numerical value is around 1.961496881, which shows that we should be quite careful with the upper bound... – asomog Mar 14 '18 at 17:47

3 Answers3

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We have \begin{align*} \frac{1}{n\sqrt{n+2}}\leq\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n+1}},\qquad n\geq 2, \end{align*} and here is why: First note that $\sqrt{n^2-1}\leq n$ for all $n\geq 1$. Moreover, $\sqrt{n+1}-\sqrt{n-1}=\frac{2}{\sqrt{n+1}+\sqrt{n-1}}\geq \frac{1}{\sqrt{n+2}}$. From this we get \begin{align*} \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n-1}}{\sqrt{n^2-1}}\geq\frac{1}{n\sqrt{n+2}}, \end{align*} as claimed.

Now, we obtain \begin{align*} \sum_{n=1}^\infty\frac{1}{n\sqrt{n+2}}&\leq\sum_{n=1}^6\frac{1}{n\sqrt{n+2}}+\sum_{n=7}^\infty\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n+1}}\right) \\ &=\frac{1}{4}+\frac{1}{\sqrt{3}}+\frac{1}{3 \sqrt{5}}+\frac{5}{4 \sqrt{6}}+\frac{6}{5 \sqrt{7}}+\frac{1}{12 \sqrt{2}}<2. \end{align*}

sranthrop
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  • Very nice, thanks! Just one more thing, how did you derive the first inequality? – Vic. Utap Mar 14 '18 at 18:03
  • Thank you, you are welcome! Honestly, it was just trial and error. I wanted to end up with a telescoping sum because then the sum is easy to calculate, and I tried several things, but I thought this was the easiest. – sranthrop Mar 14 '18 at 18:22
  • The bounding sum is approximately $1.999215$. What motivated you to use the inequality valid for $n\ge 2$ and then apply it for the sum from $7$ on? – Mark Viola Mar 14 '18 at 18:36
  • I found $1/(n\sqrt {n+2};)<2(1/\sqrt {n-1};-1/\sqrt n)$ and since the LHS and RHS are asymptotically equal, your method can be used. But I didn't calculate the arithmetic & it appears to be inferior.........+1 – DanielWainfleet Mar 14 '18 at 18:38
  • Well, since the bounding series (beginning at $n=2$) converges to $1+1/\sqrt{2}\approx 1.70711<2$, it was clear that if we let it start at some $m>2$ and fill the cap with the remaining terms of the correct series, we will eventually end up with a total value less than 2. Then I used a computer to find the smallest $m$ for which this happens, and got $m=7$. – sranthrop Mar 14 '18 at 18:42
  • As I mentioned in another comment, it was just trial and error. – sranthrop Mar 14 '18 at 18:54
  • I first used numerical experiments so see if this inequality has a chance to be true. Then I proved it analytically. I'm going to add a proof to my answer, and I hope that I didn't do anything wrong so far. – sranthrop Mar 14 '18 at 18:59
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Hint $n\sqrt{n+2}>n^{3/2}$, compare with $\int_1^{\infty}x^{-3/2}dx$.

Tsemo Aristide
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  • Thank you! But I forgot to mention that I would like to prove my inequality without using calculus. – Vic. Utap Mar 14 '18 at 17:07
  • Also, the comparison goes the wrong way: $\int_1^\infty x^{-3/2}dx$ is less than $\sum_1^\infty{1\over n^{3/2}}$, not greater than it. – Barry Cipra Mar 14 '18 at 17:11
  • It is not conclusive since $x^{-3/2}\le n^{-3/2}$. – user Mar 14 '18 at 17:12
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    This would be an upper bound for $\sum_{n=2}^\infty \cdots$. – copper.hat Mar 14 '18 at 17:34
  • This idea (and simple variations of it, such as comparing with the integral of $1/(x\sqrt{x+2})$) does not work, since it produces an estimate of the sum that is greater than two. Since the sum of the original series is greater than $1.96$, the error between the sum and the estimate must be less than $0.04$. – K B Dave Mar 14 '18 at 17:39
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Disclaimer: this is a pretty technical approach. The inverse Laplace transform of $\frac{1}{x\sqrt{x+2}}$ is given by $\frac{1}{\sqrt{2}}\text{Erf}(\sqrt{2s})$, hence the have the identity $$ \sum_{n\geq 1}\frac{1}{n\sqrt{n+2}} = \frac{1}{\sqrt{2}}\int_{0}^{+\infty}\frac{\text{Erf}(\sqrt{2s})}{e^s-1}\,ds\stackrel{s\mapsto u^2/2}{=}\frac{1}{\sqrt{2}}\int_{0}^{+\infty}\frac{u\,\text{Erf}(u)}{e^{u^2/2}-1}\,du $$ and the wanted inequality can be recovered from the continued fraction expansion of the error function and the Cauchy-Schwarz inequality.

A simpler approach is to notice that $\frac{1}{n\sqrt{n+2}}< \frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}$ and $\frac{1}{n\sqrt{n+2}}> \frac{2}{\sqrt{n+\frac{1}{6}}}-\frac{2}{\sqrt{n+\frac{7}{6}}}$ hold for any $n\geq 1$, hence by creative telescoping $$ \sum_{n\geq 1}\frac{1}{n\sqrt{n+2}}\in \frac{1}{\sqrt{3}}+\left(\sqrt{\frac{24}{13}},\sqrt{2}\right)=(1.93608271\ldots,1.99156383\ldots). $$

Jack D'Aurizio
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