Find the value of the power series sum: n=2(1)n(2n+1)(2n+2)3n ?

1 Answer
Jun 7, 2018

n=2(1)n13n1(2n+1)(2n+2)=3arctan(13)32ln(43)1736

Explanation:

Consider the geometric series:

n=0qn=11q

for |q|<1.

Let q=x2

11+x2=n=0(1)nx2n

Integrate term by term:

x0dt1+t2=n=0(1)nx0t2ndt

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

Integrate again, considering that integrating by parts:

arctanxdx=xarctanxx1+x2dx

arctanxdx=xarctanx12d(1+x2)1+x2

arctanxdx=xarctanx12ln(1+x2)+c

so:

x0arctant=n=0(1)n2n+1x0t2n+1dt

xarctanx12ln(1+x2)=n=0(1)nx2n+2(2n+1)(2n+2)

Let now x=13. As 0<13<1 the series is convergent in this point:

13arctan(13)12ln(1+13)=n=0(1)n1(3)2n+21(2n+1)(2n+2)

note that:

(3)2n+2=3(3)2n=33n

so:

13arctan(13)12ln(1+13)=13n=0(1)n13n1(2n+1)(2n+2)

Extract now the first two terms for n=0 and n=1:

13arctan(13)12ln(1+13)=161108+13n=2(1)n13n1(2n+1)(2n+2)

13arctan(13)12ln(1+13)16+1108=13n=2(1)n13n1(2n+1)(2n+2)

3arctan(13)32ln(43)1736=n=2(1)n13n1(2n+1)(2n+2)