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 Post subject: Series and inequalityPosted: Fri Jan 01, 2016 1:50 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Show that:

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

Source

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 Post subject: Re: Series and inequalityPosted: Fri Jan 01, 2016 1:51 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
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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 Post subject: Re: Series and inequalityPosted: Fri Jan 01, 2016 1:53 pm
 Team Member

Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
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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 Post subject: Re: Series and inequalityPosted: Fri Jan 01, 2016 1:54 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
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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