Friday, 1 July 2016

summation - Calculating the infinite sum $1-frac 1 7+frac 1 9 - frac{1}{15} + frac 1 {17}mp ...=frac{1+sqrt{2}}{8}pi$



Prove that

$$1-\dfrac 1 7+\dfrac 1 9 - \dfrac{1}{15} + \dfrac 1 {17}\mp ...=\dfrac{1+\sqrt{2}}{8}\pi$$



My attempt: I tried to break it into two series
$$(1+1/9+1/17+...)-(1/7+1/15+1/23+...)$$
But I don't know how to proceed. Any hints would be appreciated.


Answer



Using the hints by Mohammad Zuhair Khan and Feng Shao, let



$$f(x):=1-\sum_{n=0}^\infty\left(\frac{x^{8n-1}}{8n-1}-\frac{x^{8n+1}}{8n+1}\right).$$




Then if we differentiate term-wise,



$$f'(x)=-\sum_{n=0}^\infty(x^{8n-2}-x^{8n}).$$



Using the geometric sum formula,



$$f'(x)=-\frac{x^6}{1-x^8}+\frac{x^8}{1-x^8}=-\frac{x^6(1-x^2)}{1-x^8}.$$



Finally,




$$f(1)=1-\int_0^1\frac{x^6(1-x^2)}{1-x^8}dx.$$



https://www.wolframalpha.com/input/?i=integrate+x%5E6(1-x%5E2)%2F(1-x%5E8)+from+0+to+1



I see no easy way to solve the integral, other than by decomposition in simple fractions, which is tedious.


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...