Infinite sum

Calculus (Integrals, Series)
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Tolaso J Kos
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Infinite sum

#1

Post by Tolaso J Kos »

Evaluate the infinite sum:

$$ \sum_{n=1}^{\infty} \frac{n+3}{2^n (n+1)(n+2)} $$
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Papapetros Vaggelis
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Re: Infinite sum

#2

Post by Papapetros Vaggelis »

Use the identity \(\displaystyle{\ln\,\left(1+x\right)=\sum_{n=1}^{\infty}(-1)^{n+1}\,\dfrac{x^{n}}{n}\,\,,-1<x\leq 1\,\,(I)}\) .

The idea of the exercise here

helped me to solve this one. On the above exercise there is an extra question marked as \(\displaystyle{(\ast)}\)

For every \(\displaystyle{n\in\mathbb{N}}\) holds :

\(\displaystyle{\begin{aligned} \dfrac{n+3}{\left(n+1\right)\left(n+2\right)}&=\dfrac{\left(n+2\right)+1}{\left(n+1\right)\left(n+2\right)}\\&=\dfrac{1}{n+1}+\dfrac{1}{\left(n+1\right)\left(n+2\right)}\\&=\dfrac{1}{n+1}+\dfrac{\left(n+2\right)-\left(n+1\right)}{\left(n+1\right)\left(n+2\right)}\\&=\dfrac{1}{n+1}+\dfrac{1}{n+1}-\dfrac{1}{n+2}\\&=\dfrac{2}{n+1}-\dfrac{1}{n+2}\end{aligned}}\)

so:

\(\displaystyle{\sum_{n=1}^{\infty}\dfrac{n+3}{2^{n}\,\left(n+1\right)\,\left(n+2\right)}=\sum_{n=1}^{\infty}\left[2\,\dfrac{2^{-n}}{n+1}-\dfrac{2^{-n}}{n+2}\right]}\)

where:

\(\displaystyle{\begin{aligned} \sum_{n=1}^{\infty}\dfrac{2^{-n}}{n+1}&=\sum_{n=1}^{\infty}(-1)^{n}\,\dfrac{(-1/2)^{n}}{n+1}\\&=\sum_{n=2}^{\infty}(-1)^{n-1}\,\dfrac{(-1/2)^{n-1}}{n}\\&=-2\,\sum_{n=2}^{\infty}(-1)^{n+1}\,\dfrac{(-1/2)^{n}}{n}\\&=-2\,\left[\sum_{n=1}^{\infty}(-1)^{n+1}\,\dfrac{(-1/2)^{n}}{n}+\dfrac{1}{2}\right]\\&\stackrel{(I)}{=}-2\,\ln\,\left(1-\dfrac{1}{2}\right)-1\\&=2\,\ln\,2-1\end{aligned}}\)

and

\(\displaystyle{\begin{aligned} \sum_{n=1}^{\infty}\dfrac{2^{-n}}{n+2}&=\sum_{n=1}^{\infty}(-1)^{n}\,\dfrac{(-1/2)^{n}}{n+2}\\&=\sum_{n=3}^{\infty}(-1)^{n-2}\,\dfrac{(-1/2)^{n-2}}{n}\\&=4\,\sum_{n=3}^{\infty}(-1)^{n}\,\dfrac{(-1/2)^{n}}{n}\\&=-4\,\sum_{n=3}^{\infty}(-1)^{n+1}\,\dfrac{(-1/2)^{n}}{n}\\&=-4\,\left[\sum_{n=1}^{\infty}(-1)^{n+1}\,\dfrac{(-1/2)^{n}}{n}+\dfrac{1}{2}+\dfrac{1}{8}\right]\\&\stackrel{(I)}{=}-4\,\ln\,\left(1-\dfrac{1}{2}\right)-2-\dfrac{1}{2}\\&=4\,\ln\,2-\dfrac{5}{2}\end{aligned}}\)

Thus, the initial sereis converges and according to the algebra of limits we get :

\(\displaystyle{\sum_{n=1}^{\infty}\dfrac{n+3}{2^{n}\,\left(n+1\right)\,\left(n+2\right)}=4\,\ln\,2-2-4\,\ln\,2+\dfrac{5}{2}=\dfrac{1}{2}}\)
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