Complicated Infinite Sum

Let $f(s) = \displaystyle\sum_{n = 1}^{\infty}\frac{d(n) + \displaystyle\sum_{m = 1}^{v_2(n)}(m - 3)d\left(\frac{n}{2^m}\right)}{n^s}$ . Let $z(s) = \displaystyle\sum_{n = 1}^{\infty}\frac{1}{n^s}$ . Then,

We claim that $f(s) = z(s)^2\left(1 - \frac{1}{2^s} - \frac{1}{4^s} - \frac{1}{8^s} - ... \right)^2$ , which by the infinite geometric series formula gives $f(s) = z(s)^2\left(\frac{2^s - 2}{2^s - 1}\right)$ .

Indeed, $z(s)^2 = \displaystyle\sum_{n = 1}^{\infty}\frac{d(n)}{n^s}$ , because the value $\frac{1}{n^s}$ appears exactly $d(n)$ times in the sum.

Also, $\left(1 - \frac{1}{2^s} - \frac{1}{4^s} - \frac{1}{8^s} - ... \right)^2 = 1 + \displaystyle\sum_{i = 2}^{\infty}\frac{i - 1}{(2^i)^s} - 2\displaystyle\sum_{i = 1}^{\infty}\frac{1}{(2^i)^s} = 1 + \displaystyle\sum_{i = 1}^{\infty}\frac{i - 3}{(2^i)^s}$ . Thus, we need to prove that $f(s) = \displaystyle\sum_{n = 1}^{\infty}\frac{d(n)}{n^s} + \displaystyle\sum_{n = 1}^{\infty}\frac{d(n)}{n^s}\displaystyle\sum_{i = 1}^{\infty}\frac{i - 3}{(2^i)^s}$ .

Note that $\displaystyle\sum_{n = 1}^{\infty}\frac{d(n)}{n^s}\displaystyle\sum_{i = 1}^{\infty}\frac{i - 3}{(2^i)^s} = \displaystyle\sum_{n = 1}^{\infty}\frac{\displaystyle\sum_{m = 1}^{v_2(n)}(m - 3)d\left(\frac{n}{2^m}\right)}{n^k}$ (Split $n = 2^rs$ )

Thus, we have the desired conclusion.

So, we need to find $\lim_{s \rightarrow 1}z(s)^2(2^s - 2)$ , since $2^s - 1 = 1$ vanishes.

Note that it is well known (by Laurent Series) that $\lim_{s \rightarrow 1}(z_s - \frac{1}{s - 1}) = \gamma$ , where $\gamma$ is the Euler-Macheroni Constant. (see here , for instance). Thus, $\lim_{s \rightarrow 1}z(s)^2(2^s - 2) = \lim_{s \rightarrow 1}\frac{(2^s - 2)^2}{(s - 1)^2} + 2\gamma\lim_{s \rightarrow 1}\frac{(2^s - 2)^2}{s - 1} + \gamma^2\lim_{s \rightarrow 1}(2^s - 2)^2$ , and by L'Hopital Rule the last two terms vanish. It remains to find $\left(\lim_{s \rightarrow 1}\frac{(2^s - 2)^2}{(s - 1)^2}\right)^2$ .

By L'Hopital Rule again, $\lim_{s \rightarrow 1}\frac{2^s - 2}{s - 1} = \lim_{s \rightarrow 1}\frac{2^s \ln 2}{1} = 2 \ln 2 = \ln 4$ , so the desired sum is $(\ln 4)^2$ , which gives us $1000n + m = 2(1000) + 4 = 2004$ .