# Thread: Sum of an infinite series

1. ## Sum of an infinite series

Hi I'm stuck on a question, and sadly my coursebook is actually missing the pages I need due to a printing error.

Find the sum of the infinite series:

(5/6)^3 - (5/6)^4 + (5/6)^5 - (5/6) ^ 6 ...

Any help would be great.

2. Originally Posted by edmo
Hi I'm stuck on a question, and sadly my coursebook is actually missing the pages I need due to a printing error.

Find the sum of the infinite series:

(5/6)^3 - (5/6)^4 + (5/6)^5 - (5/6) ^ 6 ...

Any help would be great.
Start by writing the series in summation form, identify what type of series it is, and then apply the correct method to solve it.

Here is a start.
$\sum_{n=3}^{\infty}(-1)^{n+1}\bigg(\frac{5}{6}\bigg)^n$

3. Starting from the series...

$\sum_{n=0}^{\infty} (-\frac{5}{6})^{n} = 1 - \frac{5}{6} + (\frac{5}{6})^{2} - (\frac{5}{6})^{3} + \dots = \frac{1}{1+ \frac{5}{6}} = \frac{6}{11}$

... and setting...

$x= (\frac{5}{6})^{3} - (\frac{5}{6})^{4} + (\frac{5}{6})^{5} - (\frac{5}{6})^{6} + \dots$ (2)

... we arrive at the equation...

$1 - \frac{5}{6} + (\frac{5}{6})^{2} - x = \frac{6}{11}$ (3)

... the solution of which is $x=\frac{125}{396}$ ...

Kind regards

$\chi$ $\sigma$

4. Originally Posted by edmo
Hi I'm stuck on a question, and sadly my coursebook is actually missing the pages I need due to a printing error.

Find the sum of the infinite series:

(5/6)^3 - (5/6)^4 + (5/6)^5 - (5/6) ^ 6 ...

Any help would be great.
Hi edmo,

Another way is......

$\left(\frac{5}{6}\right)^3-\left(\frac{5}{6}\right)^4+\left(\frac{5}{6}\right )^5-......$

$=\left(\frac{5}{6}\right)^3\left(1-\frac{5}{6}+\left(\frac{5}{6}\right)^2-\left(\frac{5}{6}\right)^3+.....\right)$

$=\frac{125}{216}\left(1+\left(\frac{5}{6}\right)^2 +\left(\frac{5}{6}\right)^4+.......\right)-\frac{125}{216}\left(\frac{5}{6}+\left(\frac{5}{6} \right)^3+\left(\frac{5}{6}\right)^5+.....\right)$

These are 2 infinite geometric series in brackets.

$a=first\ term,\ r=geometric\ ratio,\ S_{\infty}=\frac{a}{1-r}$

For the first one.... $\frac{a}{1-r}=\frac{1}{1-\frac{25}{36}}=\frac{1}{\frac{36-25}{36}}=\frac{36}{11}$

For the 2nd one..... $\frac{a}{1-r}=\frac{\frac{5}{6}}{1-\frac{25}{36}}=\frac{5}{6}\ \frac{36}{11}=\frac{30}{11}$

Hence, the sum of all the terms is

$\frac{125}{216}\left(\frac{36}{11}-\frac{30}{11}\right)=\frac{125}{216}\ \frac{6}{11}=\frac{125}{11}\ \frac{6}{216}=\frac{125}{11}\ \frac{1}{36}$