# How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given 1/(1*3)+1/(2*4)+1/(3*5)+...+1/(n(n+2))+...?

Feb 10, 2017

The partial sum is $= \frac{3}{4} - \frac{2 n + 3}{2 \left(n + 1\right) \left(n + 2\right)}$
The series converge to $\frac{3}{4}$

#### Explanation:

Let's perform a decomposition into partial fractions

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

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

We compare the numerators

$1 = A \left(n + 2\right) + B n$

When $n = 0$, $\implies$, $1 = 2 A$, $\implies$, $A = \frac{1}{2}$

When $n = - 2$, $\implies$, $1 = - 2 B$, $\implies$, $B = - \frac{1}{2}$

Therefore,

$\frac{1}{n \left(n + 2\right)} = \frac{1}{2 n} - \frac{1}{2 \left(n + 2\right)}$

$n = 1$, $\implies$, $\frac{1}{1 \cdot 3} = \frac{1}{2} - \frac{1}{6}$

$n = 2$, $\implies$, $\frac{1}{2 \cdot 4} = \frac{1}{4} - \frac{1}{8}$

$n = 3$, $\implies$, $\frac{1}{3 \cdot 5} = \frac{1}{6} - \frac{1}{10}$

$n = 4$, $\implies$, $\frac{1}{4 \cdot 6} = \frac{1}{8} - \frac{1}{12}$

$n = 5$, $\implies$, $\frac{1}{5 \cdot 7} = \frac{1}{10} - \frac{1}{14}$

$n = n - 2$, $\implies$, $\frac{1}{\left(n - 2\right) \left(n\right)} = \frac{1}{2 \left(n - 2\right)} - \frac{1}{2 n}$

$n = n - 1$, $\implies$, $\frac{1}{\left(n - 1\right) \left(n + 1\right)} = \frac{1}{2 \left(n - 1\right)} - \frac{1}{2 \left(n + 1\right)}$

$n = n$, $\implies$, $\frac{1}{n \left(n + 2\right)} = \frac{1}{2 n} - \frac{1}{2 \left(n + 2\right)}$

Partial sum

${S}_{n} = \frac{1}{2} + \frac{1}{4} - \frac{1}{2 \left(n + 1\right)} - \frac{1}{2 \left(n + 2\right)}$

$= \frac{3}{4} - \frac{2 n + 3}{2 \left(n + 1\right) \left(n + 2\right)}$

${\lim}_{n \to + \infty} {S}_{n} = \frac{3}{4}$

The series converge to $\frac{3}{4}$

Feb 10, 2017

${\sum}_{n = 1}^{\infty} \frac{1}{n \left(n + 2\right)} = \frac{3}{4}$

#### Explanation:

We can determine that the series is convergent by direct comparison, since for $n \ge 1$:

$0 < \frac{1}{n \left(n + 2\right)} < \frac{1}{n} ^ 2$

and:

${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2 = {\pi}^{2} / 6$

is convergent.

To determine the sum we can write the general term of the series as:

$\frac{1}{n \left(n + 2\right)} = \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n + 2}\right)$

so the the partial sums are:

${s}_{N} = {\sum}_{n = 1}^{N} \frac{1}{n \left(n + 2\right)} = \frac{1}{2} {\sum}_{n = 1}^{N} \left(\frac{1}{n} - \frac{1}{n + 2}\right) = \frac{1}{2} \left(1 - \cancel{\frac{1}{3}} + \frac{1}{2} - \frac{1}{4} + \cancel{\frac{1}{3}} - \frac{1}{5} + \ldots + \frac{1}{N - 2} - \cancel{\frac{1}{N}} + \frac{1}{N - 1} - \frac{1}{N + 1} + \cancel{\frac{1}{N}} - \frac{1}{N + 2}\right)$

and we can see that in every partial sum, all the intermediate terms cancel two by two except for:

${s}_{N} = \frac{1}{2} \left(1 + \frac{1}{2} - \frac{1}{N + 1} - \frac{1}{N + 2}\right) = \frac{3}{4} - \frac{1}{2} \left(\frac{1}{N + 1} + \frac{1}{N + 2}\right)$

so that the sum of the series is:

${\sum}_{n = 1}^{\infty} \frac{1}{n \left(n + 2\right)} = {\lim}_{N \to \infty} {s}_{N} = {\lim}_{N \to \infty} \left[\frac{3}{4} - \frac{1}{2} \left(\frac{1}{N + 1} + \frac{1}{N + 2}\right)\right] = \frac{3}{4}$