Calculation of a strange series

Is it possible to find an expression for: $$S(N)=\sum_{k=0}^{+\infty}\frac{1}{\sum_{n=0}^{N}k^n}?$$

For $N=1$ we have

$$S(1) = \displaystyle\sum_{k=0}^{+\infty}\frac{1}{1 + k} = \displaystyle\sum_{k=1}^{+\infty}\frac{1}{k}$$

which is the (divergent) harmonic series. Thus, $S (1) = \infty$.

For $N=2$ this sum is: $$S(2)=\sum_{k=0}^{+\infty}\frac{1}{1+k+k^2}$$ which can be expressed as: $$S(2)=-1+\frac{1}{3}\sqrt 3 \pi \tanh(\frac{1}{2}\pi\sqrt 3)\approx 0.798$$

For $N=3$ we have: $$S(3)=\frac{1}{4}\Psi(I)+\frac{1}{4I}\Psi(I)-\frac{1}{4I}\pi\coth(\pi)+\frac{1}{4}\pi\coth(\pi)+\frac{1/}{4}\Psi(1+I)-\frac{1}{4I}\Psi(1+I)-\frac{1}{2}+\frac{1}{2}\gamma \approx 0.374$$

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The only thing I can find, using Mathematica, is that the plot of $\frac{1}{x}+1$ looks like the plot of $S_0(\text{Floor}(x))$.. – Kevin Sep 11 '12 at 12:11
I love these questions. (+1) – Chris's wise sister Sep 11 '12 at 12:17
You write "for $N = 3$", but do not introduce the symbol prior to that. What the heck is $N$? – Rod Carvalho Sep 11 '12 at 12:31
@Rod, I had the same thought. I think OP means what is later on referred to as $S_0(N)$. But it would be nice of OP to edit so it makes sense. – Gerry Myerson Sep 11 '12 at 12:45
This sum can always be rewritten, for finite $N$, as $$\sum_{k=0}^\infty\frac{1-k}{1-k^{N+1}}$$ but remains difficult as well. – Jon Sep 11 '12 at 14:21

Perform a partial fraction decomposition: $$\frac{1}{p(k)} = \frac{1}{1+k+\cdots+k^{n-1}} = \frac{1}{ \prod_{m=1}^{n-1}\left(k-\exp\left(i \frac{2 \pi}{n} m \right)\right)} = \sum_{m=1}^{n-1} \frac{1}{k-\exp\left(i \frac{2 \pi}{n} m \right)} \frac{1}{p^\prime\left(\exp\left(i \frac{2 \pi}{n} m \right)\right)}$$ Now: $$p^\prime\left(z\right) = \sum_{m=1}^{n-1} m z^{m-1} = \frac{\mathrm{d}}{\mathrm{d}z} \sum_{m=0}^{n-1} z^{m} = \frac{\mathrm{d}}{\mathrm{d}z} \frac{1-z^n}{1-z} = \frac{z-z^n (n-z(n-1))}{z (1-z)^2}$$ Therefore, using $z^n=1$ for $z=\exp\left(i \frac{2 \pi}{n} m \right)$: $$c_m := \frac{1}{p^\prime\left(\exp\left(i \frac{2 \pi}{n} m \right)\right)} = \frac{1}{n} \exp\left(i \frac{2 \pi}{n} m \right) \left( \exp\left(i \frac{2 \pi}{n} m \right) - 1 \right) = \frac{1}{n} \exp\left(i \frac{2 \pi}{n} m \right) \left( \exp\left(i \frac{2 \pi}{n} m \right) - 1 \right)$$ We thus have, and using $\sum_{m=1}^{n-1} c_m = 0$: $$\begin{eqnarray} \sum_{k=0}^\infty \frac{1}{p(k)} &=& \sum_{k=0}^\infty \sum_{m=1}^{n-1} \frac{c_m}{k-\exp\left(i \frac{2 \pi}{n} m \right)} = \sum_{k=0}^\infty \sum_{m=1}^{n-1} c_m \left(\frac{1}{k-\exp\left(i \frac{2 \pi}{n} m \right)} - \frac{1}{k+1}\right) \\ &=& -\sum_{m=1}^{n-1} c_m \sum_{k=0}^\infty \left(\frac{1}{k+1} - \frac{1}{k-\exp\left(i \frac{2 \pi}{n} m \right)}\right) \\ &=& -\sum_{m=1}^{n-1} c_m \left( \gamma + \psi\left(-\exp\left(i \frac{2 \pi}{n} m \right)\right)\right) \end{eqnarray}$$ Again, making use of $\sum_{m=1}^{n-1} c_m = 0$ we arrive at: $$\sum_{k=0}^\infty \frac{1}{1+k+\cdots+k^{n-1}} = \sum_{m=1}^{n-1} \frac{1}{n} \exp\left(i \frac{2 \pi}{n} m \right) \left(1- \exp\left(i \frac{2 \pi}{n} m \right) \right) \cdot \psi\left(-\exp\left(i \frac{2 \pi}{n} m \right)\right)$$ where $\psi(x)$ denotes the digamma function.

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Hmm, I'm 2 minutes late, and the contents are almost identical. Maybe I should erase my answer…… OTL – sos440 Sep 11 '12 at 14:57
I do not think there is any need to delete your answer. – Sasha Sep 11 '12 at 15:00
Let $p(k) = 1+k+\cdots+k^{n-1}$. Then $(1-k) p(k) = 1-k^n$, thus $n-1$ roots of $p(k)$ are also roots of $1=k^{n}$. – Sasha Sep 11 '12 at 15:02
@ Sasha: obvious, I know. I deleted my comment before you replied. – Rod Carvalho Sep 11 '12 at 15:03
@sos440 Answers are slightly different, although they use the same tactics. I thought SE's rules on possible duplication refer to exact duplicate questions, and copied answers. I do not think it pertains to a situation like this. – Sasha Sep 11 '12 at 15:13

Let $T(N) = S(N-1)$. Then

\begin{align*}T(n) &= 1 + \frac{1}{n} + \sum_{k=2}^{\infty} \frac{1}{k^{n-1}+k^{n-2}+\cdots+k+1} \\ &= 1 + \frac{1}{n} + \sum_{k=2}^{\infty} \frac{k - 1}{k^n - 1} \\ &= 1 + \frac{1}{n} + \sum_{k=2}^{\infty} \frac{1}{n} \sum_{l=1}^{n-1} \frac{\omega_l (\omega_l - 1)}{k - \omega_l} \\ &= 1 + \frac{1}{n} + \sum_{k=0}^{\infty} \frac{1}{n} \sum_{l=1}^{n-1} \frac{\omega_l (\omega_l - 1)}{k + 2 - \omega_l}, \end{align*}

where $\omega_l = \exp\left(\tfrac{2\pi l i}{n}\right)$. Since

$$\frac{1}{n} \sum_{l=0}^{n-1} \omega_l (\omega_l - 1) = 0,$$

we may write

\begin{align*}T(n) &= 1 + \frac{1}{n} + \sum_{k=0}^{\infty} \frac{1}{n} \sum_{l=1}^{n-1} \omega_l (\omega_l - 1) \left( \frac{1}{k + 2 - \omega_l} - \frac{1}{k+1} \right) \\ &= 1 + \frac{1}{n} + \frac{1}{n} \sum_{l=1}^{n-1} \omega_l (\omega_l - 1) \sum_{k=0}^{\infty} \left( \frac{1}{k + 2 - \omega_l} - \frac{1}{k+1} \right) \\ &= 1 + \frac{1}{n} - \frac{1}{n} \sum_{l=1}^{n-1} \omega_l (\omega_l - 1) \left( \gamma + \psi_0 (2 - \omega_l) \right) \\ &= 1 + \frac{1}{n} - \frac{1}{n} \sum_{l=1}^{n-1} \omega_l (\omega_l - 1) \psi_0 (2 - \omega_l). \end{align*}

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$$S(N)=1+\frac1{N+1}+\sum_{k=1}^{+\infty}\left(\zeta((N+1)k-1)-\zeta((N+1)k)\right)$$

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 This expands on @Jon's comment on the main question. – Did Sep 11 '12 at 16:05