Wikipedia gives $$\sum_{k=2}^\infty(\zeta(k)-1)=1,\quad\sum_{k=1}^\infty(\zeta(2k)-1)=\frac34,\quad\sum_{k=1}^\infty(\zeta(4k)-1)=\frac78-\frac\pi4\left(\frac{e^{2\pi}+1}{e^{2\pi}-1}\right)$$ from which we can easily find $\sum_{k=1}^\infty(\zeta(2k+1)-1)$ and $\sum_{k=1}^\infty(\zeta(4k+2)-1)$. From here it's natural to ask the following

Question: Is there a known closed form for $\sum_{k=1}^\infty(\zeta(4k+1)-1)$?

Related: Closed form for $\sum\limits_{k=1}^{\infty}\zeta(4k-2)-\zeta(4k)$ and Closed form for $\sum_{k=1}^{\infty} \zeta(2k)-\zeta(2k+1)$.


Note that we are done once we have a closed form for $\sum_{k=1}^\infty(\zeta(4k+1)-\zeta(4k+3))$. So I tried the same approach as in one of the questions above and this is what I got:
We have $$\zeta(4k+1)-\zeta(4k+3)=\sum_{n\geq2}^\infty\left(1-\frac1{n^2}\right)\frac1{n^{4k+1}}$$ Hence $$\sum_{k\geq1}\zeta(4k+1)-\zeta(4k+3)=\sum_{n\geq2}\left(1-\frac1{n^2}\right)\sum_{k\geq1}\frac1{n^{4k+1}}=\sum_{n\geq2}\left(1-\frac1{n^2}\right)\frac1{n^5}\frac1{1-\frac1{n^4}}\\=\sum_{n\geq2}\frac1{n^3+n^5}.$$

In the same way I get $$\sum_{k\geq1}\zeta(4k+1)-\zeta(4k)=-\sum_{n\geq2}\frac1{n+n^2+n^3+n^4}$$ and $$\sum_{k\geq1}\zeta(4k+1)-\zeta(4k+2)=\sum_{n\geq2}\frac1{n^2+n^3+n^4+n^5}.$$ It suffices (in fact it is equally hard) to evaluate any of these series.

  • $\begingroup$ Do you consider digamma function as closed form? If so, $$\sum_{n=1}^{\infty} (\zeta(4n+1)-1)=\frac18+\frac14 (1-2\gamma-\psi(1-i)-\psi(1+i))$$ $\endgroup$ – nospoon Sep 2 '15 at 17:01

I don't know if this will be useful, but thought that is might help. So, here we go ...

We can write the sum of interest as

$$\begin{align} \sum_{k=1}^{\infty}\left(\zeta(4k+1)-1\right)&=\sum_{k=1}^{\infty}\sum_{n=2}^{\infty}\frac{1}{n^{4k+1}} \tag 1\\\\ &=\sum_{n=2}^{\infty}\left(\frac{n}{2(n^2+1)}+\frac{1}{4(n-1)}+\frac{1}{4(n+1)}-\frac{1}{n}\right) \tag 2\\\\ &=\sum_{n=2}^{\infty}\left(\frac{n}{2(n^2+1)}-\frac{1}{2n}\right) \tag 3\\\\ &+\sum_{n=2}^{\infty}\left(\frac{1}{4(n-1)}-\frac{1}{4n}\right) \\\\ &+\sum_{n=2}^{\infty}\left(\frac{1}{4(n+1)}-\frac{1}{4n}\right) \\\\ &=\frac18-\frac12\sum_{n=2}^{\infty}\frac{1}{n(n^2+1)}\tag 4 \end{align}$$

In arriving at $(1)$, we used the series definition of the Riemann-Zeta function.

In going from $(1)$ to $(2)$, we changed the order of the series, summed the interior geometric series, and expanded in partial fractions.

In going from $(2)$ to $(3)$, we merely split the sum of three convergent series.

In arriving at $(4)$, we noted that the latter two series in $(3)$ are telescoping series whose sum is $\frac18$.

Now, this last series can be written in terms of the Digamma Function as

$$\sum_{n=2}^{\infty}\frac{1}{n(n^2+1)}=-1+\gamma+\frac12\psi^0(2-i)+\frac12\psi^0(2+i)\approx 0.17186598552401 \tag 5$$

Therefore, we have

$$\sum_{k=1}^{\infty}\left(\zeta(4k+1)-1\right)=\frac58 -\frac12 \gamma-\frac14\psi^0(2-i)-\frac14\psi^0(2+i)\approx 0.039067007237995$$


Here we show that the series in $(5)$ is indeed given by


Expanding in partial fractions yields

$$\sum_{n=2}^{\infty}\frac{1}{n(n^2+1)}=\frac12\sum_{n=2}^{\infty}\left(\frac1n-\frac{1}{n+i}\right)+\frac12\sum_{n=2}^{\infty}\left(\frac1n-\frac{1}{n-i}\right) \tag 6$$

Next, using the series definition of the digamma function, we find

$$\begin{align} \psi^0(z+1)&=-\gamma+\sum_{n=1}^{\infty}\frac{z}{n(n+z)}\\\\ &=-\gamma+\sum_{n=1}^{\infty}\left(\frac1n-\frac{1}{n+z}\right)\\\\ &-\gamma+\sum_{n=1}^{\infty}\left(\frac1n-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+1+(z-1)}\right)\\\\ &=-\gamma+1+\sum_{n=2}^{\infty}\left(\frac{1}{n}-\frac{1}{n+(z-1)}\right) \tag 7\\\\ \end{align}$$

Using $(7)$ in $(6)$, we can write

$$\begin{align} \sum_{n=2}^{\infty}\frac{1}{n(n^2+1)}&=\frac12 \left(-2\gamma+2+\psi^0(2+i)+\psi^0(2-i)\right)\\\\ &=-1+\gamma +\frac12\psi^0(2+i)+\frac12\psi^0(2-i) \end{align}$$

as was to be shown!

  • $\begingroup$ Our final results are slightly different. But it appears that the result should be approximately 0.03907. $\endgroup$ – Random Variable Sep 2 '15 at 17:56
  • $\begingroup$ Found the error!! I added in my head (mistakenly) $\frac{n}{n^2+1}-\frac1n$ and was off by a minus sign. $\endgroup$ – Mark Viola Sep 2 '15 at 18:07
  • $\begingroup$ Wolfram Alpha actually says the series diverges while at the same time showing that the partial sums are converging quickly. $\endgroup$ – Random Variable Sep 2 '15 at 18:24
  • $\begingroup$ I get on WA that the series $\sum \frac{1}{n(n^2+1)}$ converges by the comparison test. $\endgroup$ – Mark Viola Sep 2 '15 at 18:50
  • $\begingroup$ I was referring to the series $\sum_{k=1}^{\infty}\left(\zeta(4k+1)-1\right)$. $\endgroup$ – Random Variable Sep 2 '15 at 18:52

Just expanding nospoon's comment, it is well-known that $$ g(x)=\sum_{n\geq 2}\left(\zeta(n)-1\right) x^{n-1} = \frac{x}{x-1}-\psi(1-x)-\gamma\tag{1}$$ so, in order to compute series like $\sum_{n\geq 2}\left(\zeta(an+b)-1\right)$, it is enough to apply a discrete Fourier transform to the middle series and to the RHS of $(1)$. So $\sum_{n\geq 2}\left(\zeta(an+b)-1\right)$ just depends on the RHS of $(1)$ evaluated over the $a$-th roots of unity. The value of $b$ chooses if there is some simplification in the sum of digamma values or not.

  • 1
    $\begingroup$ (+1)Right. Being explicit, $\displaystyle \sum_{n\ge 1} (\zeta(4n+1)-1)=\sum_{n\ge 1} \frac{1+(-1)^n}{2} (\zeta(2n+1)-1)\\=\frac12\sum_{n\ge 1} (\zeta(2n+1)-1)+\frac12\sum_{n\ge 1} i^n\frac{1+(-1)^n}{2} (\zeta(n+1)-1)$ $\endgroup$ – nospoon Sep 2 '15 at 17:17
  • $\begingroup$ "The value of $b$ chooses if there is some simplification in the sum of digamma values or not." Can you elaborate a little bit please? Thank you. $\endgroup$ – User Jan 1 '17 at 11:38

By expressing the infinite sum as a double sum and then switching the order of summation, we get

$$\begin{align} \sum_{n=1}^{\infty} \left[\zeta(4n+1) - 1 \right] &= \sum_{n=1}^{\infty} \sum_{m=2}^{\infty} \frac{1}{m^{4n+1}} = \sum_{m=2}^{\infty} \frac{1}{m}\sum_{n=1}^{\infty} \left( \frac{1}{m^{4}} \right)^{n} \\ &= \sum_{m=2}^{\infty} \frac{1}{m} \frac{\frac{1}{m^{4}}}{1-\frac{1}{m^{4}}} = \sum_{m=2}^{\infty} \frac{1}{m} \frac{1}{m^{4}-1}. \end{align}$$

One can use the series representation $$\psi(2+z) = - \gamma +1 + \sum_{n=2}^{\infty} \frac{z}{n(n+z)}, \tag{1}$$ where $\psi(x)$ is the digamma function, to show

$$\psi(2+z) + \psi(2-z) + \psi(2+iz) + \psi(2-iz) = - 4 \gamma +4 -4z^{4} \sum_{n=2}^{\infty} \frac{1}{n(n^{4}-z^{4})}.$$

Letting $z=1$, we find

$$ \begin{align} \sum_{m=2}^{\infty} \frac{1}{m(m^{4}-1)} &= 1-\gamma - \frac{1}{4} \Big(\psi(3) + \psi(1) + \psi(2+i) + \psi(2-i) \Big) \\ &= 1-\gamma - \frac{1}{4} \left(\frac{3}{2} - \gamma - \gamma + \psi(2+i)+\psi(2-i) \right) \\ &= \frac{5}{8}-\frac{\gamma}{2} - \frac{1}{4} \Big(\psi(2-i)+\psi(2+i) \Big) \\ &\approx 0.0390670072. \end{align}$$


$(1)$ I find it easier to prove that series representation for $\psi(2+z)$ using the difference equation $$\psi(N+1+z) - \psi(z) = \sum_{n=0}^{N} \frac{1}{n+z} $$ as opposed to manipulating the series representation for $\psi(1+z)$.

$$ \begin{align} \sum_{n=2}^{N} \frac{z}{n(n+z)} &= \sum_{n=2}^{N} \frac{1}{n} - \sum_{n=2}^{N} \frac{1}{n+z} \\ &= \Big( H_{N}-1 \Big) - \Big(\psi(N+1+z) - \psi(z) - \frac{1}{z} - \frac{1}{1+z} \Big)\\ &= \Big( \psi(N+1) + \gamma -1 \Big) - \Big(\psi(N+1+z) - \psi(2+z) \Big) \end{align}$$

Letting $N \to \infty$ leads to the result since $\psi(x+a) = \ln(x) + \mathcal{O} \left(\frac{1}{x} \right)$ as $x \to \infty$.

  • $\begingroup$ That is a really cool application of the difference equation for $\psi$. But I'm not sure that it is noticeably easier than working with the series. Great addition though! $\endgroup$ – Mark Viola Sep 4 '15 at 19:31
  • $\begingroup$ @Dr.MV Thanks. It felt more natural to derive it that way. $\endgroup$ – Random Variable Sep 4 '15 at 20:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.