Sum of an infinite series

**edmo** · May 11th 2010, 12:04 PM

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

The question reads-

Find the sum of the infinite series:

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

Any help would be great.

**dwsmith** · May 11th 2010, 12:43 PM

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.

The question reads-

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

**chisigma** · May 11th 2010, 12:48 PM

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

$\chi$ $\sigma$

**Archie Meade** · May 11th 2010, 01:09 PM

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.

The question reads-

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

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

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