what value does $\sum_{n=1}^{\infty} \frac{(-8)^{n-1}}{9^n}$ converge to? we have 
$\sum_{n=1}^{\infty} \frac{(-8)^{n-1}}{9^n}$
after manipulation of the expression we are taking the sum of, we get the new expression
$\frac{((-8)^n)((-8)^{-1})}{9^n}$
After further simplification we get
$\sum_{n=1}^{\infty} (-8)^{-1}$ $\cdot$ $(\frac{-8}{9})^n$ 
This form allows us to test for convergence or divergence. It will converge IF 
$\vert{R}\vert$ < $1$ where R is the common ratio. In this case the absolute value of $\frac{-8}{9}$ is indeed less than one there fore we can apply the formula $\frac{a}{1-R}$ to find the value it converges to ($a$ here would be ${-8}^{-1}$
after evaluation i got 
$\frac{-9}{136}$
However im being told this is incorrect. I double checked my algebra and know that everything is correct up to the evaluation of the value the series converges too. Can anyone help me point out what i did wrong? My goal with this post is to understand what i did wrong and where. 
 A: You are nearly right. Note the following formula:
$$a + ar + ar^2 + \dots = \frac{a}{1-r}$$
You are however adding $ar + ar^2 + ar^3 + \dots$, and hence your result should be $$ar + ar^2 + ar^3 + \dots = \frac{a}{1-r} - a = \frac{ar}{1-r}$$
(It is the same as $a \to ar$)
So it should be $$\frac{-9}{136} - \frac{1}{-8} = \frac{1}{17}$$
A: Your reasoning is all correct, save one small mistake. The expression $\frac{a}{1-R}$ applies as follows, assuming $|R|<1$:
$$\sum_{n=\mathbf{0}}^\infty aR^n = \frac{a}{1-R}$$
Notice the bold $0$ indicating the first value of the index $n$. That is, the formula applies specifically to the sum starting with $n=0$, or equivalently when the exponent of $R$ is $0$ in the first term. In your case, you wish to evaluate
$$\sum_{n=1}^\infty (-8)^{-1} \left( \frac{-8}{9} \right)^n$$
So to apply the formula, we must factor out a power of $R$ such that the first term has $0$ as the exponent on $R$. That is, we write the above as
 $$\sum_{n=1}^\infty (-8)^{-1}\left( \frac{-8}{9} \right) \left( \frac{-8}{9} \right)^{n-1} = \sum_{n=1}^\infty \frac{1}{9} \left( \frac{-8}{9} \right)^{n-1}$$
Since the exponent on $R$ is $0$ in the first term, the formula now applies with $R=-\frac{8}{9}$ and $a=\frac{1}{9}$, giving that the sum is 
$$\frac{a}{1-R} = \frac{1}{9-(-8)} = \frac{1}{17} $$
As other answerers have found.
A: $\sum_{n=1}^{\infty} \frac{(-8)^{n-1}}{9^n} = \frac{1}{9}\sum_{n=1}^{\infty} (-\frac{8}{9})^{n-1}$ is a geometric series, so it is equal to $\frac{1}{9}\lim_{n\to \infty}\frac{1-(-\frac{8}{9})^{n+1}}{1-(-\frac{8}{9})} = \frac{1}{9}\frac{1}{\frac{17}{9}} = \frac{1}{17}$.
