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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?

Michael Hardy
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Howard P
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3 Answers3

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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}

Leucippus
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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}

Michael Hardy
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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:

$$ \vert{r}\vert=\vert{e^{c}}\vert=\bigg\vert\frac{9}{10}\bigg\vert<1.$$

Hector Blandin
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