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
limn1(2n+1)3n=0



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)=n=0(1)nx2n+12n+1



Also note that arctan(13)=π6




Therefore,



π6=n=0(1)n(13)2n+12n+1=n=063(1)n(2n+1)3n



Thus,



n=0(1)n(2n+1)3n=3π6


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