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Let $a_n = 4n(4n + 1)(4n + 2)$, show that $$\sum_{i = 1}^n \frac 1{\sqrt{a_i}} \lt \frac {\sqrt 3}6 \quad \forall n \in \mathbb{N}^+.$$

I know I need to find an upper bound for $1/\sqrt{a_n}$ but I can't see how, especially with the square root. Any hints will be appreciated!

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  • $\begingroup$ You will need quite strong inequalities to get the result as the difference between the sum and the upper bound is quite small here. We have $\sum \simeq 0.2788$ while $\sqrt{3}/6 \simeq 0.2886$ $\endgroup$ – Winther Apr 24 '16 at 15:19
  • $\begingroup$ $\sum=0.2611$wolframalpha.com/input/?i=sigma+1%2Fsqrt(4k*(4k%2B1)(4k%2B2)) $\endgroup$ – Takahiro Waki May 2 '16 at 19:37
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We may notice that: $$ \forall n\geq 1,\qquad \frac{1}{\sqrt{4n(4n+1)(4n+2)}}\leq \frac{1}{2}\left(\frac{1}{\sqrt{4n-1}}-\frac{1}{\sqrt{4n+3}}\right) $$ hence it follows that: $$ \sum_{n=1}^{N}\frac{1}{\sqrt{4n(4n+1)(4n+2)}} < \left.\frac{1}{2\sqrt{4n-1}}\right|_{n=1}=\color{red}{\frac{1}{2\sqrt{3}}}$$ as wanted. Creative telescoping wins again.

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    $\begingroup$ The first inequality is quite out of the blue, how did you come up with it ? $\endgroup$ – Gabriel Romon Apr 24 '16 at 15:33
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    $\begingroup$ @LeGrandDODOM: I wanted to approximate the main term with something like $g(n)-g(n+1)$, in order to apply creative telescoping. Since the main term behaves like $\frac{1}{n^{3/2}}$, $g(n)$ has to behave like $\frac{1}{n^{1/2}}$, so I tried something of the form $g(n)=\frac{A}{\sqrt{4n+B}}$ and got a working inequality with some trial and error. $\endgroup$ – Jack D'Aurizio Apr 24 '16 at 15:36
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    $\begingroup$ This is a pretty answer (+1). $\endgroup$ – Olivier Oloa Apr 24 '16 at 15:41
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$$ \begin{align} \frac12\sum_{k=1}^\infty\left(\frac1{\sqrt{4k-1}}-\frac1{\sqrt{4k+3}}\right) &=\sum_{k=1}^\infty\frac2{\sqrt{4k-1}\sqrt{4k+3}\left(\sqrt{4k-1}+\sqrt{4k+3}\right)}\tag{1}\\ &\ge\sum_{k=1}^\infty\frac1{\sqrt{4k-1}\sqrt{4k+3}\sqrt{4k+1}}\tag{2}\\ &\ge\sum_{k=1}^\infty\frac1{\sqrt{4k}\sqrt{4k+2}\sqrt{4k+1}}\tag{3} \end{align} $$ Explanation:
$(1)$: arithmetic
$(2)$: concavity of $\sqrt{x}$ says that $\frac{\sqrt{x}+\sqrt{y}}2\le\sqrt{\frac{x+y}2}$
$(3)$: $(4k-1)(4k+3)\le4k(4k+2)$ by expanding

This says that $$ \begin{align} \sum_{k=1}^\infty\frac1{\sqrt{4k(4k+1)(4k+2)}} &\le\frac12\sum_{k=1}^\infty\left(\frac1{\sqrt{4k-1}}-\frac1{\sqrt{4k+3}}\right)\\ &=\frac12\frac1{\sqrt3}\\ &=\frac{\sqrt3}6\tag{4} \end{align} $$


Motivation

Since the terms of the sum are $\sim\frac18k^{-3/2}$, it is often useful to consider a telescoping series where the terms are differences of something $\sim\frac14k^{-1/2}$ because such a difference is $\sim\frac18k^{-3/2}$.

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    $\begingroup$ I just realized that this is the same as Jack's answer, so I have added my motivation. $\endgroup$ – robjohn Apr 24 '16 at 15:53
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An attempt :

$$\sum_{k=2}^n \frac{1}{(4k(4k+1)(4k+2))^{1/2}}\le\sum_{k=2}^n \frac{1}{(4k)^{3/2}}\le\sum_{k=2}^\infty \frac{1}{(4k)^{3/2}}\le\int_1^\infty \frac{1}{(4x)^{3/2}}=\frac{1}{4}. $$

We only get that the sum is smaller than $\frac{1}{4}+\frac{1}{\sqrt {120}}$.

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    $\begingroup$ No, your proof is flawed... This inequality $\sum_{n=1}^\infty \frac{1}{(4n)^{3/2}}\le\int_1^\infty \frac{1}{(4x)^{3/2}}$ is wrong. Even if you fix it, the bound you get is not sharp enough. Integral method is not quite sharp enough here (you can make it work but it's tedious) $\endgroup$ – Gabriel Romon Apr 24 '16 at 15:15
  • $\begingroup$ Yes, I forgot, the term $n=1$. $\endgroup$ – mrprottolo Apr 24 '16 at 15:16
  • $\begingroup$ @LeGrandDODOM The integral method can be made to work. Just explicitly sum the first $N$ terms (say $N=5$) and then use the integral test to bound the remainding terms. It's not pretty but it does work. $\endgroup$ – Winther Apr 24 '16 at 15:21
  • $\begingroup$ @Winther $N=4$ works, but without a calculator you're not going anywhere $\endgroup$ – Gabriel Romon Apr 24 '16 at 15:22
  • $\begingroup$ I had the proof with $N=5$, but I did not find it pretty :) $\endgroup$ – Olivier Oloa Apr 24 '16 at 15:24
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you can write inequality as $2\sqrt{3}<\sqrt{4n(4n+1)(4n+2)}$ squaring we get it as $12<4n(4n+1)(4n+2)$ this the function is continuously increasing as $n$ is positive. Also $n\in N+$ this base case is $1$ so plugging in we get $120$ this $12<120$ so it's true for akin as function is monotonic and increasing

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  • $\begingroup$ I don't understand you? Can you explain it in greater detail? $\endgroup$ – Colescu Apr 24 '16 at 14:39
  • $\begingroup$ Plug in value of n as $100$ then $200$ you will see the value approaches $0$ $\endgroup$ – Archis Welankar Apr 24 '16 at 14:48
  • $\begingroup$ But my question is not the limit of it. I need to prove that inequality and I can't see how this limit will help with that? $\endgroup$ – Colescu Apr 24 '16 at 14:48
  • $\begingroup$ Wait I got my mistake let me edit $\endgroup$ – Archis Welankar Apr 24 '16 at 14:55

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