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Can you change the sum of $$ \sum_{n=1}^\infty \frac{1}{k^n} $$ where $k$ is real, into infinite product? If so how ?

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    this might help you [Euler Product](https://en.wikipedia.org/wiki/Euler_product) – 2017-02-09
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    Yes you can: a product of itself and an infinite row of 1's. Another way (which is what you are probably after): $$\sum_{n=1}^\infty \frac{1}{k^n}=\left(1+{1\over k}\right)\left(1+{1\over k^2}\right)\left(1+{1\over k^4}\right)\dots-1$$ There are many ways, most of them useless. – 2017-02-09
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    That's not a geometric series. Try https://www.encyclopediaofmath.org/index.php/Dirichlet_series – 2017-02-09
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    Yes it **is** a geometric series. Look, $k$ is fixed, the sum is over $n$. – 2017-02-09
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    Why to expand in an infinite product ? It is easy to see that $\sum_{n=1}^\infty \frac{1}{k^n}= \frac{1}{k-1}$ for $|k|>1$ – 2017-02-09

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In plenty of ways. For instance $$ \sum_{n\geq 0}\frac{1}{k^n}=\prod_{m\geq 0}\left(1+k^{-2^m}\right)\tag{1} $$ is essentially stating that the binary representation of some $n\in\mathbb{N}$ is unique.
Are you sure you did not mean Euler's product, namely $$ \zeta(k)=\sum_{n\geq 1}\frac{1}{n^k}=\prod_{p\in\mathcal{P}}\left(1-\frac{1}{p^k}\right)^{-1}\tag{2}$$ (for every $k$ such that $\text{Re}(k)>1$) ?

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    Yes it's not the zeta function , thank You very much ! – 2017-02-09