Question #ddee6

2 Answers
Apr 10, 2017

0

Explanation:

to determine the sum of a series first we need to check whether the series going to converge or diverge (I hope you understand both the terms)
out equation is =>
(-1)^n/((2n-1)(2n+1))
we see that the numerator is going to be a finite quantity as it remains 1 or -1 to the power n
but upon seeing the denominator we came across that
the term (2n-1)and (2n+1) both terms are growing to infinity
which shows us that denominator is growing to infinity
hence , c/oo=0
where is a constant and the summation of the series is =0

Apr 10, 2017

-(pi+2)/4

Explanation:

Set S=sum_{n=0}^infty(-1)^n/{(2n+3)(2n+1)}. The series converges absolutely since the denominator has degree 2 in n. So we can rearrange:

1/{(2n+3)(2n+1)}=A/(2n+3)+B/(2n+1)

So we have

A(2n+1)+B(2n+3)=1, i.e. A=-B and A+3B=1, so A=-1/2 and B=1/2

Hence

S=-1/2sum_{n=0}^infty(-1)^n/(2n+3)+1/2sum_{n=0}^infty(-1)^n/(2n+1)=-1/2sum_{n=1}^infty(-1)^{n+1}/(2n+1)+1/2sum_{n=0}^infty(-1)^n/(2n+1)=-sum_{n=1}^infty(-1)^{n+1}/(2n+1) + 1/2

Since now we have the [What is the Taylor series of f(x)=arctan(x)?]

arctan(x)=\sum_{n=0}^infty (-1)^{n+1}/(2n+1)x^{2n+1} for |x|\leq 1,
x\ne \pm i, then

pi/4=arctan(1)=\sum_{n=0}^infty (-1)^{n+1}/(2n+1)=\sum_{n=1}^infty (-1)^{n+1}/(2n+1)-1

In particular

S=-pi/4-1+1/2=-(pi+2)/4