Saturday, July 7, 2018

calculus - Finding the sum for the following series



Evaluate



n=0(1)n(2n+1)3n



I can show very easily that this series converges using the alternating series test. By setting



bn=1(2n+1)3n       bnbn+1

and
lim



However, what is the sum of the series? I can't find it. I tried to write it out term by term but I don't see any pattern.


Answer



Note that the power series of \arctan(x) is
\arctan(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}



Also note that \arctan(\frac{1}{\sqrt{3} }) = \frac{\pi}{6}




Therefore,



\frac{\pi}{6} = \sum_{n=0}^{\infty} (-1)^n \frac{\big(\frac{1}{\sqrt{3}}\big)^{2n+1}}{2n+1} = \sum_{n=0}^{\infty} \frac{6}{\sqrt{3}} \frac{(-1)^n}{(2n+1)3^n}



Thus,



\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)3^n} = \frac{\sqrt{3}\pi }{6}


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