Sum of an infinite series

May 2010
1
0
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.
 
Last edited:

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
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.
\(\displaystyle \sum_{n=3}^{\infty}(-1)^{n+1}\bigg(\frac{5}{6}\bigg)^n\)
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
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

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
Dec 2009
3,120
1,342
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......

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

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

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

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

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