# Thread: The Sum of this Series?

1. ## The Sum of this Series?

Hey all!

The series looks like this

$3 - \frac{6}{5} + \frac{12}{25} - \frac{24}{125} + \frac{48}{625} - ...$

so the first thing I did was re-wrote it as the sum, from n=1 to infinity,
$\frac{(-1)^{(n+1)}*3*2^{(n-1)}}{5^{(n-1)}}$

not sure if that is the best way to write it, and how to find the infinite sum, any hints for the right direction would be helpful! thank you.

2. Is...

$\sum_{n=0}^{\infty} (-1)^{n}\cdot x^{n} = \frac{1}{1+x}, |x|<1$

... so that...

$3\cdot \sum_{n=0}^{\infty} (-1)^{n}\cdot (\frac{2}{5})^{n} = \frac{3}{1+\frac{2}{5}} = \frac{15}{7}$

Kind regards

$\chi$ $\sigma$

3. Originally Posted by matt.qmar
Hey all!

The series looks like this

$3 - \frac{6}{5} + \frac{12}{25} - \frac{24}{125} + \frac{48}{625} - ...$

so the first thing I did was re-wrote it as the sum, from n=1 to infinity,
$\frac{(-1)^{(n+1)}*3*2^{(n-1)}}{5^{(n-1)}}$

not sure if that is the best way to write it, and how to find the infinite sum, any hints for the right direction would be helpful! thank you.

So the general element of the sum is 3*(-2/5)^(n-1). Put the 3 aside for the time being, and then we have an infinite geometric sum with first element = 1 and quotient |-2/5| < 1 so it convergest to 1/[1 -(-2/5)] = 5/7 ==> the series sum is 15/7

Tonio