<s>[INST] Calculus of $\sum_{n=1}^{\infty}( \frac{1}{2^n})$, $n$ is even I want to prove that $\sum_{n=1}^{\infty}( \frac{1}{2^n}) = \frac{1}{3}$, where $n$ can only be an even integer.
That is, how can I use linear algebra to prove that $\frac{1}{2^2} + \frac{1}{2^4} + \frac{1}{2^6} + \frac{1}{2^8} + \cdots = \frac{1}{3}$?
I heard some people saying that linear algebra help them to solve the aforementioned problem. Can anyone explain to me in detail how to get the solution $\frac{1}{3}$?
Any help is very appreciated.
 [/INST] Use the fact that the set of all even (non-zero) integers is $2\mathbb N^*$. This gives the following sum to compute
$$\sum_{k=1}^{\infty} \dfrac{1}{2^{2k}}=\sum_{k=1}^{\infty}\dfrac{1}{4^k}=\dfrac{1}{4}\dfrac{1}{1-1/4}=\dfrac{1}{4}\dfrac{4}{3}=\dfrac{1}{3}.$$
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