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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    $\begingroup$ The only thing I can find, using Mathematica, is that the plot of $\frac{1}{x}+1$ looks like the plot of $S_0(\text{Floor}(x))$.. $\endgroup$ Sep 11, 2012 at 12:11
  • $\begingroup$ I love these questions. (+1) $\endgroup$ Sep 11, 2012 at 12:17
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    $\begingroup$ You write "for $N = 3$", but do not introduce the symbol prior to that. What the heck is $N$? $\endgroup$ Sep 11, 2012 at 12:31
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    $\begingroup$ @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. $\endgroup$ Sep 11, 2012 at 12:45
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    $\begingroup$ 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. $\endgroup$
    – Jon
    Sep 11, 2012 at 14:21

4 Answers 4


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.

  • $\begingroup$ Hmm, I'm 2 minutes late, and the contents are almost identical. Maybe I should erase my answer…… OTL $\endgroup$ Sep 11, 2012 at 14:57
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    $\begingroup$ I do not think there is any need to delete your answer. $\endgroup$
    – Sasha
    Sep 11, 2012 at 15:00
  • $\begingroup$ 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 $1=k^{n}$. $\endgroup$
    – Sasha
    Sep 11, 2012 at 15:02
  • $\begingroup$ @ Sasha: obvious, I know. I deleted my comment before you replied. $\endgroup$ Sep 11, 2012 at 15:03
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    $\begingroup$ @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. $\endgroup$
    – Sasha
    Sep 11, 2012 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

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


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

  • $\begingroup$ This expands on @Jon's comment on the main question. $\endgroup$
    – Did
    Sep 11, 2012 at 16:05

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 $$S(N) = 1 + \frac{1}{N+1} + \sum_{q\ge 1} (\zeta(q(N+1)-1)-\zeta(q(N+1))).$$


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