# The sum of the series 1/(1*2) - 1/(2*3) + 1/(3*4) - .... upto infinity is equal to?

Jun 17, 2017

The sum is $= 2 \ln 2 - 1$

#### Explanation:

The general term of the series is $= {\left(- 1\right)}^{n + 1} / \left(n \left(n + 1\right)\right)$

We perform a decomposition into partial fractions

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

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

So,

$1 = A \left(n + 1\right) + B n$

When $n = 0$, $\implies$, $1 = A$

When $n = - 1$, $\implies$, $1 = - B$

Therefore,

$\frac{1}{n \left(n + 1\right)} = \frac{1}{n} - \frac{1}{n + 1}$

${\left(- 1\right)}^{n + 1} / \left(n \left(n + 1\right)\right) = {\left(- 1\right)}^{n + 1} / n - {\left(- 1\right)}^{n + 1} / \left(n + 1\right)$

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

$\ln \left(1 + x\right) = {\sum}_{1}^{\infty} {\left(- 1\right)}^{n + 1} / n \cdot {x}^{n}$

${\sum}_{1}^{\infty} {\left(- 1\right)}^{n + 1} / n = \ln 2$

${\sum}_{0}^{\infty} {\left(- 1\right)}^{n + 1} / \left(n + 1\right) = {\sum}_{0}^{1} {\left(- 1\right)}^{n + 1} / \left(n + 1\right) - {\sum}_{1}^{\infty} {\left(- 1\right)}^{n} {x}^{n + 1} / \left(n + 1\right)$

${\sum}_{0}^{\infty} {\left(- 1\right)}^{n} {x}^{n + 1} / \left(n + 1\right) = 1 - \ln \left(1 + x\right)$

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

${\sum}_{1}^{\infty} {\left(- 1\right)}^{n + 1} / \left(n \left(n + 1\right)\right) = \ln 2 - \left(1 - \ln 2\right) = 2 \ln 2 - 1$