0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This site uses MathJax formatting $\endgroup$ – JKnecht Mar 16 '16 at 0:13
  • 1
    $\begingroup$ Looks to me a geometric series with a rate equal to $-0.25$. There is a formula to calculate that sum. $\endgroup$ – imranfat Mar 16 '16 at 0:15
0
$\begingroup$

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

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

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

$\endgroup$
0
$\begingroup$

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: $$S_\infty=\frac{a}{1-r}=\frac{2}{1-\frac{-1}{4}}=\frac{8}{5}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.