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#