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Series and inequality

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Tolaso J Kos
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Series and inequality

#1

Post by Tolaso J Kos » Fri Jan 01, 2016 1:50 pm

Show that:

$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} (n+1)}<2$$
Source
IMC 2015/ 2nd Round/ 1st Problem
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Papapetros Vaggelis
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Re: Series and inequality

#2

Post by Papapetros Vaggelis » Fri Jan 01, 2016 1:51 pm

Hello everybody.

We have that \(\displaystyle{0<\dfrac{1}{\sqrt{n}\,(n+1)}\leq \dfrac{1}{n\,\sqrt{n}}\,,n\in\mathbb{N}}\) and thus :

\(\displaystyle{\lim_{n\to +\infty}\dfrac{1}{\sqrt{n}\,(n+1)}=0}\) and the series converges since

\(\displaystyle{\sum_{n=1}^{\infty}\dfrac{1}{n\,\sqrt{n}}=\sum_{n=1}^{\infty}\dfrac{1}{n^{3/2}}<\infty}\).

Also, for every \(\displaystyle{n\in\mathbb{N}}\) holds:

\(\displaystyle{\begin{aligned} \dfrac{1}{(n+1)\,\sqrt{n}}&=\dfrac{1}{\sqrt{n}\,\sqrt{n+1}}\cdot \dfrac{1}{\sqrt{n+1}}\\&<\dfrac{1}{\sqrt{n}\,\sqrt{n+1}}\,\dfrac{2}{\sqrt{n}+\sqrt{n+1}}\\&=2\,\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\,\sqrt{n+1}}\\&=2\,\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\end{aligned}}\)

where

\(\displaystyle{\begin{aligned} 2\,\sum_{n=1}^{\infty}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)&=2\,\lim_{n\to +\infty}\,\sum_{k=1}^{n}\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\\&=2\,\lim_{n\to +\infty}\,\left[\left(1-\dfrac{1}{\sqrt{2}}\right)+\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\right)+...+\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\right]\\&=2\,\lim_{n\to +\infty}\left(1-\dfrac{1}{\sqrt{n+1}}\right)\\&=2\end{aligned}}\)

Therefore, \(\displaystyle{\sum_{n=1}^{\infty}\dfrac{1}{\sqrt{n}\,(n+1)}<2}\) .
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Re: Series and inequality

#3

Post by Papapetros Vaggelis » Fri Jan 01, 2016 1:53 pm

Note

We define \(\displaystyle{f:\left[1,+\infty\right)\longrightarrow \mathbb{R}}\) by \(\displaystyle{f(x)=\dfrac{1}{\sqrt{x}\,(x+1)}}\) .

The function \(\displaystyle{f}\) is continuous, positive and strictly decreasing at \(\displaystyle{\left[1,+\infty\right)}\) with

\(\displaystyle{f(1)=\dfrac{1}{2}}\) . We have that :

\(\displaystyle{\begin{aligned} \int_{1}^{+\infty}f(x)\,\mathrm{d}x&=\int_{1}^{\infty}\dfrac{1}{\sqrt{x}\,(x+1)}\,\mathrm{d}x\\&\stackrel{y=\sqrt{x}}{=}\int_{1}^{+\infty}\dfrac{2}{1+y^2}\,\mathrm{d}y\\&=\left[2\,\arctan\,y\right]_{1}^{+\infty}\\&=\pi-\dfrac{\pi}{2}\\&=\dfrac{\pi}{2}\end{aligned}}\)

According to \(\displaystyle{\rm{Cauchy's}}\) criterion, the real sequence \(\displaystyle{a_{n}=\sum_{k=1}^{n}f(k)-\int_{1}^{n}f(x)\,\mathrm{d}x}\)

converges to \(\displaystyle{\left[0,f(1)\right]=\left[0,\dfrac{1}{2}\right]}\). So,

\(\displaystyle{\begin{aligned} 0\leq \lim_{n\to +\infty}a_{n}\leq \dfrac{1}{2}&\implies 0\leq \sum_{n=1}^{\infty}\dfrac{1}{\sqrt{n}\,(n+1)}-\int_{1}^{+\infty}f(x)\,\mathrm{d}x\leq \dfrac{1}{2}\\&\implies \dfrac{\pi}{2}\leq \sum_{n=1}^{\infty}\dfrac{1}{\sqrt{n}\,(n+1)}\leq \dfrac{\pi+1}{2}\end{aligned}}\)

We deduce that \(\displaystyle{\sum_{n=1}^{\infty}\dfrac{1}{\sqrt{n}\,(n+1)}\in\left[\dfrac{\pi}{2},2\right)}\) .
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Re: Series and inequality

#4

Post by Tolaso J Kos » Fri Jan 01, 2016 1:54 pm

Here is another solution to the problem:

Let \( S \) denote the given sum, then:

$$\begin{aligned}
S=\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}\left ( n+1 \right )}&= \sum_{n=1}^{\infty}\frac{\sqrt{n}}{n(n+1)} \\
&=\sum_{n=1}^{\infty}\left [ \frac{\sqrt{n}}{n}- \frac{\sqrt{n}}{n+1} \right ]\\
&= \frac{\sqrt{1}}{1}+ \sum_{n=1}^{\infty} \left[ \frac{\sqrt{n+1}}{n+1}-\frac{\sqrt{n}}{n+1} \right]\\
&=1+ \sum_{n=1}^{\infty}\frac{1}{\left ( n+1 \right )\left ( \sqrt{n}+\sqrt{n+1} \right )} \\
&\overset{CS}{\leq} 1+ \frac{1}{4}\sum_{n=1}^{\infty}\left [ \frac{1}{\sqrt{n+1}(n+1)} + \frac{1}{\sqrt{n}(n+1)} \right ]\\
&= 1+ \frac{S}{4}+ \frac{1}{4}\sum_{n=2}^{\infty}\frac{1}{n^{3/2}} < 1+\frac{S}{4} +\frac{1}{4} \int_1^\infty \frac{{\rm d}x}{x^{3/2}}
\end{aligned}$$

and the result follows.
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