# Is my understanding of the solution for $a$ in $\sum^\infty_{n=0} e^{na} = 2$ correct?

This was a homework problem that I got and I was required to find the value of $a$ in $$\sum^\infty_{n=0} e^{na} = 2$$ So I did the following. I removed the $\sum$ and along with the indexer $n$ and was left with $e^a = 2$ where the solution to $a$ is easy to find ($a = \ln 2$). Then I know that $e \gt 2$ then the value has to be decreasing so I added a negative to $\ln 2$ which gave me the answer ($-\ln 2$). I also noticed that this the solution gave the familiar series

$$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \ldots$$ which equals to $2$ so I know that my answer is correct but what is the actual/systematic way of solving series like this?

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Call $e^a = x$. We then have $$\sum_{n=0}^{\infty} x^n = 2 \implies \dfrac1{1-x} = 2 \implies x = \dfrac12 \implies a = - \ln(2)$$ Note that $x = e^a = e^{-\ln(2)} = \dfrac12$. Hence, $$\sum_{n=0}^{\infty} e^{na} = \sum_{n=0}^{\infty} \dfrac1{2^n} = 2$$
@user17762 When you say that $\sum_{n=0}^{\infty} x^n = 2 \implies \dfrac1{1-x} = 2$, do you know that from a experience or just through the definition or something? – Jeel Shah Jun 2 '13 at 16:45