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.
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.