# How do you find a formula for the nth partial sum of the series [5/1*2]+[5/2*3]+[5/3*4]+...+[5/n(n+1)]+... and use it to find the series' sum if the series converges?

Feb 28, 2015

First of all we can factor the $5$:

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

This series, except for the factos $5$, is named Mengoli Series.

Now let's try to separate the term 1/(n(n+1) in the sum of two terms:

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

=(n(A+B)+B)/(n(n+1).

For the identity principle of polynomials:

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

So:

$A + B = 0$
$A = 1$

That gives:

$A = 1$
$B = - 1$.

So:

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

Let's see now ${S}_{n}$:

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

We can notice now that all the terms except the first and the last will erase themselves, so:

${S}_{n} = 5 \left(1 - \frac{1}{n + 1}\right)$ and we know that we can obtain the sum of ALL the terms of the series with this limit:

${\lim}_{n \rightarrow + \infty} {S}_{n} = {\lim}_{n \rightarrow + \infty} 5 \left(1 - \frac{1}{n + 1}\right) = 5 \left(1 - 0\right) = 5$.