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I want to prove $$\sum_{n=0}^{\infty }\frac{1}{(2n+1)(2n+2)}=\sum_{n=0}^{\infty }\frac{4}{(4n+1)(4n+2)(4n+3)}$$

amWhy
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E.H.E
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2 Answers2

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\begin{eqnarray} \sum_{n=0}^{\infty}\frac{1}{(2n+1)(2n+2)} &=&\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1} =\sum_{n=0}^{\infty}(\frac{1}{4n+1}-\frac{1}{4n+2}+\frac{1}{4n+3}-\frac{1}{4n+4})\\ &=&\sum_{n=0}^{\infty}(\frac{1}{4n+1}-\frac{2}{4n+2}+\frac{1}{4n+3}) +\sum_{n=0}^{\infty}(\frac{1}{4n+2}-\frac{1}{4n+4})\\ &=&\sum_{n=0}^{\infty}\frac{2}{(4n+1)(4n+2)(4n+3)} +\frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{(2n+1)(2n+2)} \end{eqnarray}

Alfred Chern
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  • Seconde equality: you can't do it without justification. $$\sum^\infty (x_n+y_n)=\sum^\infty x_n+\sum^\infty y_n$$ if and only if $\sum^\infty x_n$ and $\sum^\infty y_n$ converge. – idm Nov 09 '14 at 12:28
  • @idm: Yes, the series are convergent obviously in this case. – Alfred Chern Nov 09 '14 at 12:30
  • just precise it in your original answer and it's fine ;-) – idm Nov 09 '14 at 12:32
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$$\frac{1}{(2n+1)(2n+2)}=\frac{1}{2n+1}-\frac{1}{2n+2},$$ hence: $$\sum_{n=0}^{+\infty}\frac{1}{(2n+1)(2n+2)}=\sum_{n\geq 1}\frac{(-1)^{n-1}}{n}=\int_{0}^{1}\sum_{n\geq 0}(-x)^{n}\,dx=\int_{0}^{1}\frac{dx}{1+x}=\log 2,$$ while: $$\frac{4}{(4n+1)(4n+2)(4n+3)}=\frac{2}{4n+1}-\frac{2}{2n+1}+\frac{2}{4n+3}$$ so: $$\begin{eqnarray*}\sum_{n\geq 0}\frac{4}{(4n+1)(4n+2)(4n+3)}&=&2\int_{0}^{1}\sum_{n\geq 0}(x^{4n}-2x^{4n+1}+x^{4n+2})\,dx\\&=&2\int_{0}^{1}\frac{(1-x)^2}{1-x^4}\,dx\\&=&2\int_{0}^{1}\frac{1-x}{(1+x)(1+x^2)}\,dx\end{eqnarray*}$$ and by setting $x=\frac{1-y}{1+y}$ we get that the last integral equals: $$ \int_{0}^{1}\frac{2y}{y^2+1}\,dy=\log 2.$$

Jack D'Aurizio
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