# Thread: Error in infinite series

1. ## Error in infinite series

The question is to estimate the magnitude of error involved in using the sum of the first four terms to approximate the sum of the entire series. In this case, the series is sigma from n=1 to infinity of ((-1)^(n+1))(1/(10^n))
I'm not quite sure what im supposed to be doing here, any insight would be much appreciated! Am I finding the sum of the first four, then comparing it to the total sum to infinity?

2. The sum of the first 4 is easy enough. Just plug them in and we get .0909

The sum of the entire series you could break it up into the negative and positive.

$\left(\frac{1}{10}-\frac{1}{100}+\frac{1}{1000}-\frac{1}{10,000}+\frac{1}{100,000}+......\right)$

$\sum_{k=0}^{\infty}\frac{1}{10^{2k+1}}=\left(\frac {1}{10}+\frac{1}{1000}+\frac{1}{100,000}+.....\rig ht)=\frac{1}{10}\sum_{k=0}^{\infty}\frac{1}{100^{k }}=\frac{1}{10}\cdot\frac{1}{1-\frac{1}{100}}=\frac{10}{99}$

$\sum_{k=1}^{\infty}\frac{-1}{100^{k}}\left(\frac{-1}{100}-\frac{1}{10,000}-\frac{1}{100,000}-....\right)=\frac{\frac{-1}{100}}{1-\frac{1}{100}}=\frac{-1}{99}$

$\frac{10}{99}-\frac{1}{99}=\frac{1}{11}$

3. $\sum\limits_{n\, = \,1}^\infty {\frac{{( - 1)^{n + 1} }}
{{10^n }}} = \frac{1}
{{10}}\sum\limits_{n\, = \,0}^\infty {\frac{{( - 1)^n }}
{{10^n }}} = \frac{1}
{{10}} \cdot \frac{1}
{{1 + \dfrac{1}
{{10}}}} = \frac{1}
{{11}}.$