# Sum of an infinite series series

I've determined the convergence of this alternating series. It converges absolutely. The series is $$\sum_{i=1}^\infty 2\left(\frac{-1}{4}\right)^{(n-1)}.$$ I'm stuck on how to find the sum of this series. It could be a mental block but I just can't think of a way to determine the sum.

• This site uses MathJax formatting – JKnecht Mar 16 '16 at 0:13
• Looks to me a geometric series with a rate equal to $-0.25$. There is a formula to calculate that sum. – imranfat Mar 16 '16 at 0:15

$$\sum_{n=1}^\infty \left(-\frac{1}{4}\right)^{n-1} = \left. \sum_{n=1}^\infty x^{n-1} \right|_{-1/4} = \left. \frac{1}{1-x} \right|_{-1/4} = \frac{4}{5}$$ Since ($|x| < 1$) $$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$

• There is a $2$ upfront... – imranfat Mar 16 '16 at 0:17
• @imranfat I think OP can figure that out easily. – Henricus V. Mar 16 '16 at 0:17
• That is also true... – imranfat Mar 16 '16 at 0:18

Hint : $$\sum_{i=0}^n q^i = \frac{1-q^{n+1}}{1-q}$$

The other submitter had the general idea. The series $$\sum_{n=1}^\infty 2\left(\frac{-1}{4}\right)^{n-1}$$ is a geometric series of the form $$\sum_{n=1}^\infty {a{r}^{n-1}}=a+ar+ar^2+ar^3+...$$ where the first term is $a$ and the common ratio is $r$. This geometric series always converges if $\left|r\right|\lt 1$, where in this case the sum is given by: $$S_\infty=\frac{a}{1-r}$$

Comparing coefficients, it can be seen that $a=2$ and $r=\frac{-1}{4}$.

Therefore, since $\left|r\right|=\frac{1}{4}\lt 1$, the sum of the series is given by: $$S_\infty=\frac{a}{1-r}=\frac{2}{1-\frac{-1}{4}}=\frac{8}{5}$$