Question #ddee6

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 $\implies$
${\left(- 1\right)}^{n} / \left(\left(2 n - 1\right) \left(2 n + 1\right)\right)$
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 $\left(2 n - 1\right)$and $\left(2 n + 1\right)$ both terms are growing to infinity
which shows us that denominator is growing to infinity
hence , $\frac{c}{\infty} = 0$
where is a constant and the summation of the series is =0

Apr 10, 2017

$- \frac{\pi + 2}{4}$

Explanation:

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

$\frac{1}{\left(2 n + 3\right) \left(2 n + 1\right)} = \frac{A}{2 n + 3} + \frac{B}{2 n + 1}$

So we have

$A \left(2 n + 1\right) + B \left(2 n + 3\right) = 1$, i.e. $A = - B$ and $A + 3 B = 1$, so $A = - \frac{1}{2}$ and $B = \frac{1}{2}$

Hence

$S = - \frac{1}{2} {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(2 n + 3\right) + \frac{1}{2} {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(2 n + 1\right) = - \frac{1}{2} {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / \left(2 n + 1\right) + \frac{1}{2} {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(2 n + 1\right) = - {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / \left(2 n + 1\right) + \frac{1}{2}$

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

$\arctan \left(x\right) = \setminus {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n + 1} / \left(2 n + 1\right) {x}^{2 n + 1}$ for $| x | \setminus \le q 1$,
$x \setminus \ne \setminus \pm i$, then

$\frac{\pi}{4} = \arctan \left(1\right) = \setminus {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n + 1} / \left(2 n + 1\right) = \setminus {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / \left(2 n + 1\right) - 1$

In particular

$S = - \frac{\pi}{4} - 1 + \frac{1}{2} = - \frac{\pi + 2}{4}$