Evaluation of series

Calculus (Integrals, Series)
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Tolaso J Kos
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Evaluation of series

#1

Post by Tolaso J Kos »

Evaluate the series:

$$\sum _{n=0}^{\infty }\frac{1}{(4n+1)(4n+3)}$$
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Papapetros Vaggelis
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Re: Evaluation of series

#2

Post by Papapetros Vaggelis »

Firstly, the series \(\displaystyle{\sum_{n=0}^{\infty}\dfrac{1}{\left(4\,n+1\right)\,\left(4\,n+3\right)}}\) converges, since

\(\displaystyle{4\,n+1>n+1\,\,,4\,n+3>n+1\,,\forall\,n\in\mathbb{N}\cup\left\{0\right\}}\) and thus :

\(\displaystyle{\dfrac{1}{(4\,n+1)\,(4\,n+3)}<\dfrac{1}{(n+1)^2}}\) and

\(\displaystyle{\sum_{n=0}^{\infty}\dfrac{1}{(n+1)^2}=\sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}<\infty}\).

Therefore, \(\displaystyle{0\leq \sum_{n=0}^{\infty}\dfrac{1}{(4\,n+1)\,(4\,n+3)}\leq \dfrac{\pi^2}{6}}\) .

If \(\displaystyle{n\in\mathbb{N}\cup\left\{0\right\}}\), then :

\(\displaystyle{\begin{aligned} a_{n}&=\dfrac{1}{\left(4\,n+1\right)\,\left(4\,n+3\right)}\\&=\dfrac{1}{2}\,\dfrac{\left(4\,n+3\right)-\left(4\,n+1\right)}{\left(4\,n+1\right)\,\left(4\,n+3\right)}\\&=\dfrac{1}{2}\,\dfrac{1}{4\,n+1}-\dfrac{1}{2}\,\dfrac{1}{4\,n+3}\\&=\dfrac{1}{2}\,\int_{0}^{1}x^{4\,n}\,\mathrm{d}x-\dfrac{1}{2}\,\int_{0}^{1}x^{4\,n+2}\,\mathrm{d}x\\&=\int_{0}^{1}x^{4\,n}\,\left(\dfrac{1}{2}-\dfrac{x^2}{2}\right)\,\mathrm{d}x\end{aligned}}\)

So, \(\displaystyle{\sum_{n=0}^{\infty}\dfrac{1}{\left(4\,n+1\right)\,\left(4\,n+3\right)}=\sum_{n=0}^{\infty}\,\int_{0}^{1}x^{4\,n}\,\left(\dfrac{1}{2}-\dfrac{x^2}{2}\right)\,\mathrm{d}x}\) .

Let \(\displaystyle{\left(f_{n}\right)_{n\in\mathbb{N}\cup\left\{0\right\}}}\) be the sequence of real fucntions

\(\displaystyle{f_{n}:\left[0,1\right]\longrightarrow \mathbb{R}\,,x\mapsto f_{n}(x)=x^{4\,n}\,\left(\dfrac{1}{2}-\dfrac{x^2}{2}\right)}\) .

If \(\displaystyle{x\in\left[0,1\right]}\), then: \(\displaystyle{f_{n}(x)\longrightarrow 0\,,n\to\infty}\) and for each

\(\displaystyle{x\in\left[0,1\right)}\) and for every \(\displaystyle{n\in\mathbb{N}\cup\left\{0\right\}}\) holds :

\(\displaystyle{\left|f_{n}(x)\right|=(x^4)^{n}\,\left(\dfrac{1}{2}-\dfrac{x^2}{2}\right)\leq \dfrac{(x^2)^{n}}{2}}\)

and \(\displaystyle{\sum_{n=0}^{\infty}(x^2)^n=\dfrac{1}{1-x^2}<\infty\,,x\in\left[0,1\right)}\), which means that the function \(\displaystyle{\left(f_{n}\right)}\)

converges uniformly on \(\displaystyle{\left[0,1\right)}\).

What about point \(\displaystyle{x=1}\) ? In order to overcome this singularity, we use a limit procession :

Since the polynomial functions are smooth functions, we have that :

\(\displaystyle{\int_{0}^{1}x^{4\,n}\,\left(\dfrac{1}{2}+\dfrac{x^2}{2}\right)\,\mathrm{d}x=\lim_{x\to 1^{-}}\,\int_{0}^{x}t^{4\,n}\,\left(\dfrac{1}{2}-\dfrac{t^2}{2}\right)\,\mathrm{d}t}\) .

Now,

\(\displaystyle{\sum_{n=0}^{\infty}\dfrac{1}{\left(4\,n+1\right)\,\left(4\,n+3\right)}=\sum_{n=0}^{\infty}\,\lim_{x\to 1^{-}}\int_{0}^{x}t^{4\,n}\,\left(\dfrac{1}{2}-\dfrac{t^2}{2}\right)\,\mathrm{d}t\,\,(I)}\)

Consider the sequence \(\displaystyle{\left(g_{n}\right)_{n\in\mathbb{N}\cup\left\{0\right\}}}\) of real functions

\(\displaystyle{g_{n}:\left[0,1\right)\longrightarrow \mathbb{R}\,,x\mapsto g_{n}(x):=\int_{0}^{x}f_{n}(t)\,\mathrm{d}t}\) .

Obviously, \(\displaystyle{g_{n}(x)\geq 0\,,\forall\,x\in\left[0,1\right)\,,\forall\,n\in\mathbb{N}\cup\left\{0\right\}}\) and

\(\displaystyle{0\leq g_{n}(x)\leq \int_{0}^{x}(t^2)^{n}\,\mathrm{d}t=(x^2)^{n+1}\,\dfrac{1}{n+1}}\)

where \(\displaystyle{\lim_{n\to +\infty}(x^2)^{n+1}\,\dfrac{1}{n+1}=0\,,\forall\,x\in\left[0,1\right)}\), so :

\(\displaystyle{\lim_{n\to +\infty}g_{n}(x)=0\,,\forall\,x\in\left[0,1\right)}\) . Also,

\(\displaystyle{\begin{aligned} \sum_{n=0}^{\infty}\dfrac{(x^2)^{n+1}}{n+1}&=\sum_{n=1}^{\infty}\dfrac{(x^2)^{n}}{n}\\&=-\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}\,(-x^2)^{n}}{n}\\&\stackrel{-1<-x^2\leq 0}{=}-\ln\,\left(1-x^2\right)<\infty\end{aligned}}\)

(and \(\displaystyle{-\ln\,\left(1-x^2\right)>0}\), since \(\displaystyle{0<1-x^2\leq 1}\) ).

Therefore, the sequence \(\displaystyle{\left(g_{n}\right)_{n\in\mathbb{N}\cup\left\{0\right\}}}\) converges uniformly and thus :

\(\displaystyle{\begin{aligned} \sum_{n=0}^{\infty}\dfrac{1}{\left(4\,n+1\right)\,\left(4\,n+3\right)}&\stackrel{(I)}{=}\sum_{n=0}^{\infty}\lim_{x\to 1^{-}}g_{n}(x)\\&=\lim_{x\to 1^{-}}\sum_{n=0}^{\infty}g_{n}(x)\\&=\lim_{x\to 1^{-}}\sum_{n=0}^{\infty}\int_{0}^{x}t^{4\,n}\,\left(\dfrac{1}{2}-\dfrac{t^2}{2}\right)\,\mathrm{d}t\\&=\lim_{x\to 1^{-}}\,\int_{0}^{x}\,\sum_{n=0}^{\infty}\dfrac{1-t^2}{2}\,(t^4)^{n}\\&=\lim_{x\to 1^{-}}\,\int_{0}^{x}\,\dfrac{1-t^2}{2}\,\dfrac{1}{1-t^4}\\&=\lim_{x\to 1^{-}}\,\int_{0}^{x}\,\dfrac{1-t^2}{2}\,\dfrac{1}{\left(1-t^2\right)\,\left(1+t^2\right)}\\&=\lim_{x\to 1^{-}}\,\int_{0}^{x}\dfrac{1}{2\,\left(1+t^2\right)}\,\mathrm{d}t\\&=\lim_{x\to 1^{-}}\left[\dfrac{\arctan\,t}{2}\right]_{0}^{x}\\&=\lim_{x\to 1^{-}}\dfrac{\arctan\,x}{2}\\&=\dfrac{1}{2}\,\dfrac{\pi}{4}\\&=\dfrac{\pi}{8}\in\left(0,\dfrac{\pi^2}{6}\right)\end{aligned}}\) .
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Riemann
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Re: Evaluation of series

#3

Post by Riemann »

Another way would be using digamma. Here is how:

\begin{align*}
\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+1 \right )\left ( 4n+3 \right )} &= \frac{1}{2}\sum_{n=0}^{\infty}\left [ \frac{1}{4n+1}- \frac{1}{4n+3} \right ]\\
&= \frac{1}{8}\sum_{n=0}^{\infty}\left [ \frac{1}{n+1/4}- \frac{1}{n+3/4} \right ]\\
&= \frac{1}{8}\left [ \psi \left ( \frac{3}{4} \right )- \psi \left ( \frac{1}{4} \right ) \right ]\\
&\overset{(*)}{=} \frac{\pi\cot \frac{\pi}{4}}{8}= \frac{\pi}{8}
\end{align*}

$(*)$ Due to the digamma reflection formula $\psi\left ( 1-z \right )-\psi(z)= \pi \cot \pi z $.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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