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How can I find the Dirichlet generating function of $C_n^2$ $n\geq1$ and $2^\frac{n}{d}$ , $n\geq1$ , $d|n$.

I tried a lot of time to do both of these , setting the series in the formula : $\displaystyle \sum_{n=1}^{\infty}\frac{a_n}{n^s}$ , but I did not get anywhere.

I'll be so grateful if you can help me.

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  • $\begingroup$ ${n \choose 2}=c_0+c_1n+c_2n^2$ thus $\sum_n {n \choose 2}n^{-s}=...$. For the other sequence check the convergence first. $\endgroup$ – reuns Dec 3 '19 at 14:27
  • $\begingroup$ What are $c_0 ,c_1 ,c_2$ ? $\endgroup$ – Almaa Dec 3 '19 at 15:53
  • $\begingroup$ What is your guess, do you know the definition of ${n \choose 2}$ $\endgroup$ – reuns Dec 3 '19 at 16:07
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    $\begingroup$ ${n \choose 2} = \frac{n(n-1)}{2} = \frac12 n^2- \frac12 n$ thus $\sum_{n\ge 1} {n \choose 2}n^{-s}=\frac12 \zeta(s-2)-\frac12 \zeta(s-1)$ for $s > 3$. For the other series I told you to check the convergence. Does $2^{n/d}n^{-s}\to 0$ ? $\endgroup$ – reuns Dec 4 '19 at 19:23
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    $\begingroup$ If you prefer formal Dirichlet series then $Li_s(z) = \sum_{n\ge 1} z^n n^{-s}$, usually we consider $|z|< 1$ but you can put $z=2$ to get a formal Dirichlet series, the point is that (for $z\ne \pm$) $f(n)=z^n$ is not multiplicative thus it is not quite natural nor useful to look at its Dirichlet generating function. $\endgroup$ – reuns Dec 4 '19 at 19:36
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Dirichlet series for $C_n^2$

$$C_n^2 = \frac{n(n-1)}{2} = \frac{n^2 - n}{2}$$

$$\sum_{n = 1}^\infty \frac{C_n^2}{n^s} = \frac{1}{2}(\sum_{n = 1}^\infty \frac{n^2}{n^s} - \sum_{n = 1}^\infty \frac{n}{n^s}) = \frac{\zeta(s - 2)-\zeta(s - 1)}{2}$$

Here $\zeta$ stands for Riemann zeta function, which is the Dirichlet generating function for the sequence $\{1\}_{n = 1}^\infty$. It is well known and well studied, but, unfortunately, non-elementary. However, if you like, it can be expressed as $\zeta(s) = \frac{\int_0^\infty \frac{x^{s - 1}}{e^x - 1}dx}{\int_0^\infty x^{s - 1}e^{-x}dx}$.

Dirichlet series for $2^\frac{n}{d}$

Suppose, $n = dk$. Then $\sum_{n = 1}^\infty \frac{2^\frac{n}{d}}{n^s} = \frac{1}{d^s}\sum_{k = 1}^\infty \frac{2^k}{k^s}$. Thus it will be sufficient for us to find the Dirichlet generating function for $2^n$. And here it is.

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