Question

Question #ddee6

Answer 1

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

Answer 2

`-(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`