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.

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- May 11th 2010, 12:04 PMedmoSum 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.

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. - May 11th 2010, 12:43 PMdwsmith
- May 11th 2010, 12:48 PMchisigma
Starting from the series...

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

... and setting...

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

... we arrive at the equation...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - May 11th 2010, 01:09 PMArchie Meade
Hi edmo,

Another way is......

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

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

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

These are 2 infinite geometric series in brackets.

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

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

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

Hence, the sum of all the terms is

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