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

Apr 4, 2018

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

#### Explanation:

Given the series:

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

we can determine that is convergent using the direct comparison test since:

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

and the series:

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

is convergent based on the $p$-series test.

Decompose now the general term using partial fractions:

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

and note that in the partial sum:

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

all terms cancel each other except the first and the last, so:

${s}_{n} = 1 - \frac{1}{n + 1}$

and then:

${\lim}_{n \to \infty} {s}_{n} = 1$

is the sum of the series.

Apr 4, 2018

The answer is $= 1$

#### Explanation:

${\sum}_{k = 1}^{n} \frac{1}{k \left(k + 1\right)} = {\sum}_{k = 1}^{n} \frac{1}{k} - {\sum}_{k = 1}^{n} \frac{1}{k + 1}$

$= {\sum}_{k = 1}^{n} \frac{1}{k} - {\sum}_{k = 2}^{n + 1} \frac{1}{k}$

$= 1 + \cancel{{\sum}_{k = 2}^{n} \frac{1}{k}} - \cancel{{\sum}_{k = 2}^{n} \frac{1}{k}} - \frac{1}{n + 1}$

$= 1 - \frac{1}{n + 1}$

$= \frac{n}{n + 1}$

${\lim}_{n \to + \infty} \frac{n}{n + 1} = {\lim}_{n \to + \infty} \frac{1}{1 + \frac{1}{n}} = 1$

As

${\lim}_{n \to + \infty} \frac{1}{n} = 0$