# How do I write (5/(1*2))+(5/(2*3))+(5/(3*4))+...+(5/n(n+1))+...in summation notation, and how can I tell if the series converges?

$5 \setminus {\sum}_{k = 1}^{n} \setminus \frac{1}{k \left(k + 1\right)}$

#### Explanation:

The given series:

$\left(\frac{5}{1 \setminus \cdot 2}\right) + \left(\frac{5}{2 \setminus \cdot 3}\right) + \left(\frac{5}{3 \setminus \cdot 4}\right) + \setminus \ldots + \left(\frac{5}{n \left(n + 1\right)}\right)$

$= \setminus {\sum}_{k = 1}^{n} \setminus \frac{5}{k \left(k + 1\right)}$

$= 5 \setminus {\sum}_{k = 1}^{n} \setminus \frac{1}{k \left(k + 1\right)}$

$= 5 \setminus {\sum}_{k = 1}^{n} \left(\frac{1}{k} - \setminus \frac{1}{k + 1}\right)$

$= 5 \left(\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \setminus \ldots + \left(\frac{1}{n} - \frac{1}{n + 1}\right)\right)$

$= 5 \left(1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \setminus \ldots + \frac{1}{n} - \frac{1}{n + 1}\right)$

$= 5 \left(1 - \frac{1}{n + 1}\right)$

$\setminus \therefore \setminus {\lim}_{n \setminus \to \setminus \infty} \setminus {\sum}_{k = 1}^{n} \setminus \frac{5}{k \left(k + 1\right)}$

$= \setminus {\lim}_{n \setminus \to \setminus \infty} 5 \left(1 - \frac{1}{n + 1}\right)$

$= 5 \left(1 - 0\right)$

$= 5$

Hence, the given series is converging

Jul 19, 2018

Please see some comlementary details below.

#### Explanation:

I added a few more details

The partial fraction decomposition is

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

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

Compare the numerators

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

Let $k = 0$, $\implies$, $1 = A$

Let $k = - 1$, $\implies$, $1 = - B$

Therefore,

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

Therefore,

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

$= 1 + {\sum}_{k = 2}^{n} \frac{1}{k} - \left({\sum}_{k = 1}^{n - 1} \frac{1}{k + 1}\right) + \frac{1}{n + 1}$

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

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

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

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

And

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

$= 1$

The series converges and the limit is $= 5$