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:
$$S(3)=\frac{1}{4}\Psi(I)+\frac{1}{4I}\Psi(I)-\frac{1}{4I}\pi\coth(\pi)+\frac{1}{4}\pi\coth(\pi)+\frac{1/}{4}\Psi(1+I)-\frac{1}{4I}\Psi(1+I)-\frac{1}{2}+\frac{1}{2}\gamma \approx 0.374$$
 A: 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.
A: 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
$$ \begin{align*}T(n) 
&= 1 + \frac{1}{n} + \sum_{k=0}^{\infty} \frac{1}{n} \sum_{l=1}^{n-1} \omega_l (\omega_l - 1) \left( \frac{1}{k + 2 - \omega_l} - \frac{1}{k+1} \right) \\
&= 1 + \frac{1}{n} + \frac{1}{n} \sum_{l=1}^{n-1} \omega_l (\omega_l - 1) \sum_{k=0}^{\infty} \left( \frac{1}{k + 2 - \omega_l} - \frac{1}{k+1} \right) \\
&= 1 + \frac{1}{n} - \frac{1}{n} \sum_{l=1}^{n-1} \omega_l (\omega_l - 1) \left( \gamma + \psi_0 (2 - \omega_l) \right) \\
&= 1 + \frac{1}{n} - \frac{1}{n} \sum_{l=1}^{n-1} \omega_l (\omega_l - 1) \psi_0 (2 - \omega_l).
\end{align*}$$
A: $$
S(N)=1+\frac1{N+1}+\sum_{k=1}^{+\infty}\left(\zeta((N+1)k-1)-\zeta((N+1)k)\right)
$$
A: Here is an  approach using Mellin transforms to  enrich the collection
of solutions. Write
$$S(N) = 1 + \frac{1}{N+1} + \sum_{k\ge 2} \frac{1}{\sum_{n=0}^N k^n}
= 1 + \frac{1}{N+1} + \sum_{k\ge 2} \frac{k-1}{k^{N+1}-1}
\\= 1 + \frac{1}{N+1} + \sum_{k\ge 2} \frac{k}{k^{N+1}-1}
- \sum_{k\ge 2} \frac{1}{k^{N+1}-1}.$$
There are two harmonic sums with  the same base function here which we
now evaluate.
Introduce $$S_1(x, M) = \sum_{k\ge 2} \frac{1}{(xk)^M-1}
\quad\text{and}\quad S_2(x, M) = \sum_{k\ge 2} \frac{k}{(xk)^M-1}$$
so that we are interested in $S_2(1, N+1)-S_1(1, N+1).$
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have for $k\ge 2$ that 
(do $S_1$ first as $S_2$ will follow)
$$\lambda_k = 1, \quad \mu_k = k 
\quad \text{and} \quad
g(x) = \frac{1}{x^M-1}.$$
We need the Mellin transform $g^*(s)$ of $g(x)$ which is
$$\int_0^\infty \frac{1}{x^M-1} x^{s-1} dx.$$
We take $M =  N+1 > 2$ since the sum from  the beginning diverges when
$N=1.$

The  fundamental strip  of this Mellin transform is
$\langle 0,M\rangle.$ We will in fact use $\langle 0, M-1\rangle$

Now to  evaluate this (seemingly  divergent) transform we use  a slice
contour with  the bottom side of  the slice aligned  with the positive
real axis  and the  origin. The  angle of the  slice is  $2\pi/M.$ The
radius of the slice is $R$ and we let it go to infinity so that we can
easily  see  the contribution  from  the  arc  that connects  the  two
straight  sides  vanishes  by  the  ML bound  because  its  length  is
$\Theta(R)$ and  the integrand is  $$\Theta(1/R^M\times R^{(M-1)-1}) =
\Theta(1/R^2).$$

We will  be integrating  through the poles  at $x=1$  and $x=\exp(2\pi
i/M),$  thereby picking  up half  the residues  from these  poles. The
integral along  the horizontal  side $\Gamma_1$ of  the slice  is our
transform $g^*(s)$.  The contribution  from the arc call it $\Gamma_2$
vanishes.   Along   the  slanted  side  call  it   $\Gamma_3$  we  put
$x=\exp(2\pi i/M)t$ obtain the following integral:
$$\int_{\Gamma_3} \frac{1}{x^M-1} x^{s-1} dx
\\= \exp(2\pi i/M)
\int_\infty^0 \frac{1}{(\exp(2\pi i/M) t)^M-1} 
(\exp(2\pi i/M) t)^{s-1} dt
\\ = - \exp(2\pi i/M)
\int_0^\infty \frac{1}{t^M-1} 
(\exp(2\pi i/M) t)^{s-1} dt
\\= - \exp(2\pi i \times s/M)
\int_0^\infty \frac{1}{t^M-1} t^{s-1} dt =
- \exp(2\pi i \times s/M) \times g^*(s).$$
Putting it all together we have for $g^*(s)$ that
$$g^*(s) = \left(1-e^{2\pi i \times s/M}\right)
= \pi i
\left(
\mathrm{Res}\left(\frac{1}{x^M-1} x^{s-1}; x=1\right)+
\mathrm{Res}\left(\frac{1}{x^M-1} x^{s-1}; x=\exp(2\pi i/M)\right)
\right).$$
These poles are  simple and may be evaluated  with a single derivative
which gives $1/(M x^{M-1})$ and produces
$$ \pi i 
\left(\frac{1}{M} + 
\frac{1}{M \exp(2\pi i/M)^{M-1}} 
\exp(2\pi i \times (s-1)/M)\right).$$
This is
$$\frac{\pi i}{M}
\left(1 + \exp(2\pi i \times (s-1-(M-1))/M)\right)
\\= \frac{\pi i}{M}
\left(1 + \exp(2\pi i \times (s-M)/M)\right)
\\= \frac{\pi i}{M}
\left(1 + \exp(2\pi i \times s/M)\right).$$
Returning to $g^*(s)$ we finally obtain
$$g^*(s) = \frac{\pi i}{M}
\frac{1 + e^{2\pi i \times s/M}}{1-e^{2\pi i \times s/M}}
= \frac{\pi}{M}
\frac{i(e^{-\pi i \times s/M} + e^{\pi i \times s/M})}
{e^{-\pi i \times s/M}-e^{\pi i \times s/M}}
= - \frac{\pi}{M} \cot(\pi s/M).$$
It  follows that  the Mellin  transform $Q_1(s)$  of the  harmonic sum
$S_1(x,M)$ is given by
$$Q_1(s) = - \frac{\pi}{M} \cot(\pi s/M) (-1+\zeta(s))
\\ \text{because}\\
\sum_{k\ge 2} \frac{\lambda_k}{\mu_k^s} = 
\sum_{k\ge 2} \frac{1}{k^s}
= -1+\zeta(s)$$
for $\Re(s) > 1.$
Similarly  the Mellin  transform $Q_2(s)$  of the  harmonic sum
$S_2(x,M)$ is given by
$$Q_2(s) = - \frac{\pi}{M} \cot(\pi s/M) (-1+\zeta(s-1))
\\ \text{because}\\
\sum_{k\ge 2} \frac{\lambda_k}{\mu_k^s} = 
\sum_{k\ge 2} k \times \frac{1}{k^s}
= -1+\zeta(s-1)$$
for $\Re(s) > 2.$
The Mellin inversion integral for the first one is
$$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q_1(s)/x^s ds$$
which we evaluate  by shifting it to the right  for an expansion about
infinity. For the second one we get
$$\frac{1}{2\pi i} \int_{5/2-i\infty}^{5/2+i\infty} Q_2(s)/x^s ds$$
Now  note  that  the first  pole  to  the  right  of the  abscissa  of
convergence is at $s=M$ and since $M\ge 3>5/2>3/2$ we may in fact join
these  two inversion integrals  and write
$$\frac{1}{2\pi i} \int_{5/2-i\infty}^{5/2+i\infty} 
(Q_2(s)-Q_1(s))/x^s ds$$
which is 
(the minus sign disappears because we are integrating clockwise in the
right half-plane)
$$\frac{1}{2\pi i} \int_{5/2-i\infty}^{5/2+i\infty} 
\frac{\pi}{M} \cot(\pi s/M) (\zeta(s-1)-\zeta(s))/x^s ds.$$
Observe that
$$\mathrm{Res}\left(\cot(\pi s/M); s=qM\right) = \frac{M}{\pi}$$
with $q$ an integer. 
Collecting the  residues at the  poles at $s  = qM$ in the  right half
plane and setting $x=1$ we obtain the convergent series for
$S_2(1, M)-S_1(1, M)$
$$\sum_{q\ge 1} \frac{\pi}{M} \frac{M}{\pi} (\zeta(qM-1)-\zeta(qM))
= \sum_{q\ge 1} (\zeta(qM-1)-\zeta(qM)).$$
Returning  to $N$  and the  sum  we started  with we  can confirm  the
earlier result that
$$S(N) = 1 + \frac{1}{N+1} 
+ \sum_{q\ge 1} (\zeta(q(N+1)-1)-\zeta(q(N+1))).$$
