Calculating the exponent of a series I'm trying to solve for the variable $c$ in the equation
$$\sum_{n=0}^\infty e^{nc} = 10$$
At first I've recognized this is a geometric series, and I know that if $\vert{r}\vert < 1$ the value of this series can be expressed by:
$$\sum_{n=0}^\infty ar^{n} = \lim_{n\to\infty}S_{n} = \lim_{n\to\infty}\frac{a}{1-r}(1 - r^n)=\frac{a}{1-r}$$
since when $\vert{r}\vert\geq 1$ the geometric series will diverge and in the contrary case will converge because:
$$ \vert{r}\vert < 1 \Rightarrow \lim_{n\to\infty}r^{n}=0 $$
In my case this means $r=e^{c}$ and if we suppose that $\vert{e^{c}}\vert<1$ we get
$$ \frac{1}{1-e^{c}}=\lim_{n\to\infty}\frac{1}{1-e^{c}}  (1 - e^{nc}) = 10$$
I try some basic algebraic manipulation
$$  1 - e^{nc} = 10 - 10e$$
To simplify:
$$e^{nc} = -9 + 10e$$
$$\ln(e^{nc}) = \ln(-9 + 10e)$$
$$nc = \ln(-9 + 10e)$$
$$c = \frac{\ln(-9 + 10e)}{n}$$
Now this doesn't feel entirely right, and I wanted to verify the answer/if I made any mistakes. The answer shouldn't be an equation but just a number, where did I mess up?
 A: Using 
$$\sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$$
then 
$$ 10 = \sum_{n=0}^{\infty} e^{c \, n} = \frac{1}{1 - e^{c}}$$
From this it is then determined that
\begin{align}
1 - e^{c} &= \frac{1}{10} \\
e^{c} &= \frac{9}{10} \\
c &= 2 \, \ln 3 - \ln10.
\end{align}
A: You have $\dfrac 1 {1-e^c} = 10,$ but then after that you introduced a strange complication, bringing the index $n$ into it. The index $n$ is equal to $0$ in the first term, to $1$ in the second term, to $2$ in the third term, and so on, but what is  it supposed to be equal to in the equation that says $1-e^{nc} = \text{something?}$ What you need to do is just go with what you had before that:
\begin{align}
\frac 1 {1-e^c} & = 10 \\[10pt]
1 - e^c & = \frac 1 {10} \\[10pt]
e^c & = \frac 9 {10} \\[10pt]
c & = \ln \frac 9 {10}
\end{align}
A: In general you can use that for any integer $s\geq 0$ we have
$$ \sum_{n=s}^{\infty}r^{n} =\frac{r^{s}}{1-r} $$
if $\ \vert{r}\vert<1$. Also remark that in your case it's ok because from the previous answer you know that:
$$ \vert{r}\vert=\vert{e^{c}}\vert=\bigg\vert\frac{9}{10}\bigg\vert<1.$$
