Concerning this curious sum: $\sum_{n=1}^{\infty}\frac{2n+1}{2^{2n}}\left[\frac{n-1}{2^{2n}}-(n-2)\right]\zeta(2n+1)=\ln2$

Question

How would you go about showing this is correct? $$\sum_{n=1}^{\infty}\frac{2n+1}{2^{2n}}\left[\frac{n-1}{2^{2n}}-(n-2)\right]\zeta(2n+1)=\ln2$$

Any hints may help?

It is quite starnge $\zeta(2n+1)$ would cough out $\ln2$

Answer 1

Hint.

One may start with the following identiy (https://dlmf.nist.gov/5.7): $$ \sum _{k=1}^{\infty } (-1)^k\zeta (k+1)\:x^k=-\psi(x+1)-\gamma,\quad |x|<1, $$ where $\psi(x)=\left(\ln \Gamma(x)\right)' $ giving, by differentiating twice, an identity involving $$ \sum _{n=2}^{\infty }(2n+1)(n-1)\zeta (2 n+1) x^{2 n}, $$ then put $x=\frac12$ and $x=\frac14$ to get the result.

Answer 2

Mathematica seems to "know" the result. It spits out the sum in less than a second

$$\sum _{n=1}^{\infty } \frac{(2 n+1) \left(\frac{n-1}{2^{2 n}}-(n-2)\right) \zeta (2 n+1)}{2^{2 n}}= \log(2)\tag{1}$$

What does it do internally? Inserting the definition of the Zeta function

$$\zeta(2n+1) = \sum_{k=1}^{\infty} \frac{1}{k^{2n+1}}$$

into the left hand side of $(1)$ and exchanging the order of summation we find for the $n$-sum the strange expression

$$\begin{align} \sum _{n=1}^{\infty } \frac{(2 n+1)}{2^{2 n} k^{2 n+1}} \left(\frac{n-1}{2^{2 n}}-(n-2)\right) \\ = \frac{196608 k^{10}-179200 k^8+34240 k^6-2304 k^4+40 k^2-1}{k \left(4 k^2-1\right)^3 \left(16 k^2-1\right)^3} \end{align}\tag{2}$$

This doesn't look simple at all, but - surspise! - the remaining $k$-sum results in $$\sum _{k=1}^{\infty } \frac{196608 k^{10}-179200 k^8+34240 k^6-2304 k^4+40 k^2-1}{k (2 k-1)^3 (2 k+1)^3 (4 k-1)^3 (4 k+1)^3} = \log (2)\tag{3}$$

which - beside $(1)$ - is one of the most peculiar representation of $\log(2)$

Why is it so?

The partial fraction decomposition is

$$\begin{align} \frac{1}{k (2 k-1)^3 (2 k+1)^3 (4 k-1)^3 (4 k+1)^3}=\frac{47}{81 (2 k-1)}-\frac{7}{72 (2 k-1)^2}+\frac{1}{108 (2 k-1)^3}+\frac{47}{81 (2 k+1)}+\frac{7}{72 (2 k+1)^2}+\frac{1}{108 (2 k+1)^3} -\frac{256}{81 (4 k-1)}+\frac{16}{27 (4 k-1)^2}-\frac{32}{27 (4 k-1)^3}-\frac{256}{81 (4 k+1)}-\frac{16}{27 (4 k+1)^2}-\frac{32}{27 (4 k+1)^3}+\frac{1}{k} \end{align}\tag{4} $$

This doesn't help much because I can't spot any cancellations.

The partial sum is

$$\begin{align}\sum _{k=1}^m \frac{196608 k^{10}-179200 k^8+34240 k^6-2304 k^4+40 k^2-1}{k (2 k-1)^3 (2 k+1)^3 (4 k-1)^3 (4 k+1)^3} =\frac{1}{8 (2 m+1)^3}(-64 \gamma m^3-128 m^3 \psi ^{(0)}\left(\frac{1}{2}\right)+64 m^3 \psi ^{(0)}\left(\frac{3}{2}\right)-64 m^3 \psi ^{(0)}(m+1)+128 m^3 \psi ^{(0)}\left(m+\frac{1}{2}\right)-64 m^3 \psi ^{(0)}\left(2 m+\frac{3}{2}\right)+8 m^3 \psi ^{(2)}\left(\frac{1}{2}\right)-8 m^3 \psi ^{(2)}\left(\frac{3}{2}\right)-8 m^3 \psi ^{(2)}\left(m+\frac{1}{2}\right)+8 m^3 \psi ^{(2)}\left(2 m+\frac{3}{2}\right)-96 \gamma m^2+64 m^2-192 m^2 \psi ^{(0)}\left(\frac{1}{2}\right)+96 m^2 \psi ^{(0)}\left(\frac{3}{2}\right)-96 m^2 \psi ^{(0)}(m+1)+192 m^2 \psi ^{(0)}\left(m+\frac{1}{2}\right)-96 m^2 \psi ^{(0)}\left(2 m+\frac{3}{2}\right)+12 m^2 \psi ^{(2)}\left(\frac{1}{2}\right)-12 m^2 \psi ^{(2)}\left(\frac{3}{2}\right)-12 m^2 \psi ^{(2)}\left(m+\frac{1}{2}\right)+12 m^2 \psi ^{(2)}\left(2 m+\frac{3}{2}\right)-48 \gamma m+48 m-96 m \psi ^{(0)}\left(\frac{1}{2}\right)+48 m \psi ^{(0)}\left(\frac{3}{2}\right)-48 m \psi ^{(0)}(m+1)+96 m \psi ^{(0)}\left(m+\frac{1}{2}\right)-48 m \psi ^{(0)}\left(2 m+\frac{3}{2}\right)+6 m \psi ^{(2)}\left(\frac{1}{2}\right)-6 m \psi ^{(2)}\left(\frac{3}{2}\right)-6 m \psi ^{(2)}\left(m+\frac{1}{2}\right)+6 m \psi ^{(2)}\left(2 m+\frac{3}{2}\right)-8 \psi ^{(0)}(m+1)+16 \psi ^{(0)}\left(m+\frac{1}{2}\right)-8 \psi ^{(0)}\left(2 m+\frac{3}{2}\right)-\psi ^{(2)}\left(m+\frac{1}{2}\right)+\psi ^{(2)}\left(2 m+\frac{3}{2}\right)-8 \gamma -16 \psi ^{(0)}\left(\frac{1}{2}\right)+8 \psi ^{(0)}\left(\frac{3}{2}\right)+\psi ^{(2)}\left(\frac{1}{2}\right)-\psi ^{(2)}\left(\frac{3}{2}\right)\end{align}\tag{5}$$

For $m\to \infty$ the numerator is

$$4 (2 m+1) (8 m (m+1) \log (2)-3+\log (4))\tag{6}$$

Hence the conplete expression becomes in that limit

$$\begin{align}\frac{4 (2 m+1) (8 m (m+1) \log (2)-3+\log (4))}{8 (2 m+1)^3} = \\ \frac{4 m^2 \log (2)}{(2 m+1)^2}-\frac{3}{2 (2 m+1)^2}+\frac{4 m \log (2)}{(2 m+1)^2}+\frac{\log (4)}{2 (2 m+1)^2}\to \log(2)\end{align} \tag{7}$$

Summarizing, I would be really very much interested who invented this fomula and in which way?

Answer 3

This solution is combined with some general considerations from my attempt to understand the solution of Olivier Oloa

Generating function

Let the general question be, given a "reasonable" function $f(n)$ how do we calculate a sum of the type

$$s=\sum _{n=2}^{\infty } f(n) \zeta (n)\tag{1}$$

Notice that the sum starts at $n=2$ because $\zeta(0)$ and $\zeta(1)$ have divergent series.

The natural starting point is the generating sum

$$s(z) = \sum_{n=2}^{\infty} z^n \zeta(n)\tag{2}$$

from which we can generate positive powers of $n$ by differentiating, negative ones by integrating, and exponential terms by chosing appropriate values of $z$.

$s(z)$ can easily be calculated:

$$\begin{align}s(z) = \sum_{n=2}^{\infty} z^n \zeta(n)=\sum _{n=2}^{\infty } z^n \sum _{k=1}^{\infty } \frac{1}{k^n}=\sum _{k=1}^{\infty } \left(\sum _{n=2}^{\infty } \frac{z^n}{k^n}\right)\\ =\sum _{k=1}^{\infty } \frac{z^2}{k (k-z)}=-z \sum _{k=1}^{\infty } \left(\frac{1}{k}-\frac{1}{k-z}\right)=-z H_{-z}\end{align}\tag{3}$$

where $H_x$ is the harmonic number for the real variable $x$.

Even and odd powers can be split considering the two sums

$$s_{\pm}(z) = \sum_{n=2}^{\infty} (\pm z)^n \zeta(n)= (\mp z) H_{\mp z}$$

so that for even and odd sums we find, respectively,

$$s_e(z) = \frac{1}{2}(s_{+} + s_{-}) = \sum_{k \ge 1} z^{2k}\zeta(2k)= \frac{z}{2}\left(H_z - H_{-z} \right)=\frac{1}{2} (1-\pi z \cot (\pi z))\tag{4}$$

$$s_o(z) = \frac{1}{2}(s_{+} - s_{-})=\sum_{k \ge 1} z^{2k+1}\zeta(2k+1)= -\frac{z}{2}\left(H_z + H_{-z} \right)\tag{5}$$

Applications

Examples:

Letting $z=\frac{1}{2}$ we get $s(\frac{1}{2}) = (-\frac{1}{2}) H_{-\frac{1}{2}}=(-\frac{1}{2})(-\log(4)) = \log(2)$

Remember that $H_x = \int_{0}^{1}\frac{1-t^x}{1-t}\;dt$ which leads to $H_{-\frac{1}{2}}= \int_{0}^{1}\frac{1-t^x}{1-t}\;dt\overset{t \to u^2}=-2 \int_{0}^{1}\frac{1}{1+u}\;du= -2 \log(2)$

Hence

$$\sum_{n=2}^{\infty} \frac{1}{2^n}\zeta(n)=\log(2)\tag {6a}$$

Similarly,

$$\sum _{n=2}^{\infty } \frac{n \zeta (n)}{2^n}=\frac{\partial \left(-z H_{-z}\right)}{\partial z}|_{z\to\frac{1}{2}}=\frac{\pi ^2}{8}+\log (2)\tag{6b}$$

and

$$\sum _{n=2}^{\infty } \frac{\zeta (n)}{n 2^n}= \int_0^{\frac{1}{2}} \left(-H_{-z} \right)\, dz=\frac{1}{2} (\log (\pi )-\gamma )\tag{6c}$$

Series of the OP

We wish to show that

$$\sigma :=\sum _{k\ge 1} \frac{2 k+1}{2^{2 k}} \left[\frac{k-1}{2^{2 k}}-(k-2)\right] \zeta (2 k+1)= \log(2)\tag{7}$$

by generating the sum from derivatives of $s_o$.

Rearranging we can write

$$\sigma = \sigma_1 + \sigma_2$$

with

$$\sigma_1 = - \sum_{k\ge 1} (2k+1)\zeta(2k+1) \left(\frac{1}{4^{2k}} -\frac{2}{4^{k}} \right)=\sigma_{11}+\sigma_{12}\tag{8a}$$

$$\sigma_2 = \sum_{k\ge 1} (2k+1)\;\color{blue}{(k)}\;\zeta(2k+1) \left(\frac{1}{4^{2k}} -\frac{1}{4^{k}} \right)=\sigma_{21}+\sigma_{22}\tag{8b}$$

The derivatives are, together with their values at $z=\frac{1}{2}$ and $z=\frac{1}{4}$

$$s_0(z) = \sum_{k\ge 1} \zeta(2k+1) z^{2k+1}= -\frac {z}{2}(H_z + H_{-z})\tag{9a}$$ $$s_o(z=\{\frac{1}{2}, \frac{1}{4}\})=\left\{\log (2)-\frac{1}{2},\frac{1}{4} (\log (8)-2)\right\}\tag{9b}$$

$$s_0'(z) = \sum_{k\ge 1} (2k+1)\zeta(2k+1) z^{2k}=\frac{1}{2} \left(-H_{-z}-H_z+z \left(H_z^{(2)}-H_{-z}^{(2)}\right)\right)\tag{10a}$$ $$s_o'(z=\{\frac{1}{2}, \frac{1}{4}\})=\{\log (4),\log (8)-2 C\}\tag{10b}$$

$$\begin{align}s_0''(z) = \sum_{k\ge 1} (2k+1)(2k)\zeta(2k+1) z^{2k-1}=\frac{2}{z} \sum_{k\ge 1} (2k+1)\color{blue}{(k)}\zeta(2k+1) z^{2k}\\ =z (\zeta (3,1-z)+\zeta (3,z+1))+\psi ^{(1)}(1-z)-\psi ^{(1)}(z+1)\end{align}\tag{11a}$$

$$s_o''(z=\{\frac{1}{2}, \frac{1}{4}\})=\{7 \zeta (3),14 \zeta (3)-16 C\}\tag {11b}$$

Here $C = \sum_{k\ge 0} \frac{(-1)^k}{(2k+1)^2}$ is Catalan's constant, $\psi (z)=\frac{\Gamma '(z)}{\Gamma (z)}$ is the digamma function, and $\zeta (s,a)=\sum_{k\ge 0} (a+k)^{-s}$ ist the Hurwitz-Zeta function.

Now we can identify the sums and give their values

$$\sigma_{11} = -s_0'(\frac{1}{4})=-\log (8)+2 C \tag{12a}$$ $$\sigma_{12}=2 s_0'(\frac{1}{2})=2 \log (4)\tag{12b}$$ $$\sigma_{21} = \frac{z}{2} s_o''(z)_{z\to \frac{1}{4}}=\frac{1}{8}s_o''(\frac{1}{4})=\frac{1}{8}(14 \zeta (3)-16 C)\tag{12c}$$

$$\sigma_{22} = - \frac{z}{2} s_o''(z)|_{z\to \frac{1}{2}}=-\frac{1}{4}s_o''(\frac{1}{2})=-\frac{1}{4}(7 \zeta (3))\tag{12d}$$

Notice the "miraculous" cancelleation of the more complicated expressions appearing in $(10)$ and $(11)$

Now, as an additional "miracle", in the total sum the terms with $\zeta(3)$ and $C$ cancel out leaving just

$$\sigma=\log(2)$$

which proves $(7)$.

Discussion

§1. The rather lenghty developments have provided a lot of tools to tackle similar problems.

§2. The beneficial cancellations are, of course no accident, but are due to the carefully chosen coefficients of the derivatives of $s_o(z)$.

The author of the OP has proposed in a comment a sum for Catalan's constant (cf. (a2) below) which inspired me to study these extractions more closely.

Appropriate linear combinations of the partial sums $\sigma_{ik}$ defined in $(8)$ can lead to interesting representations of known constants.

Indeed, writing the sum as

$$s=\sum_{i,j=1}^{2}c_{ij}\sigma_{ij} = q_1 C + q_2 \log(2) + q_3 \zeta(3)$$

we have after inserting the $\sigma_{ij}$ from $(12)$

$$q_1 = 2(c_{11}-c_{21}), q_2 = 4 c_{12}-3 c_{11}, q_3 = \frac{7}{4}(c_{21}-c_{22})$$

Solving these latter equations with respect to the $c_{ij}$, normalizing the sum of the $c_{ij}$ to unity, gives in terms of the $q_i$

$$\begin{array}{l} c(1,1)=\frac{4 \text{q1}}{15}-\frac{\text{q2}}{15}+\frac{16 \text{q3}}{105}+\frac{4}{15},\\ c(1,2)=\frac{\text{q1}}{5}+\frac{\text{q2}}{5}+\frac{4 \text{q3}}{35}+\frac{1}{5},\\ c(2,1)= -\frac{7 \text{q1}}{30}-\frac{\text{q2}}{15}+\frac{16 \text{q3}}{105}+\frac{4}{15},\\ c(2,2)= -\frac{7 \text{q1}}{30}-\frac{\text{q2}}{15}-\frac{44 \text{q3}}{105}+\frac{4}{15} \\ \end{array}$$

these have be put back to the sum $s$.

Now choosing the $q_i$ appropriately, e.g. $q_1=1$, $q_2=q_3=0$ we get results like the following:

$$\begin{align} s_{C,1}=C= \frac{1}{15}\sum _{n=1}^{\infty }\frac{2 n+1}{2^{2 n+1}}\left[\frac{n-16}{2^{2 n}}-(n-24)\right]\zeta(2n+1)\end {align} \tag{a1}$$

actually, $s_{C,1}=\frac{1}{15} (15 C+12 \log (4)-8 \log (8))$

The following sum was provided by the author of the OP in a comment.

$$\begin{align} s_{C,2}=C=\sum _{n=1}^{\infty }\frac{2 n+1}{2^{2 n+1}} \left[\frac{n-2}{2^{2 n}}-(n-3)\right]\zeta (2 n+1) \end {align} \tag{a2}$$

actually, $s_{C,2}=\frac{1}{2} (2 C+3 \log (4)-2 \log (8))$

Here's the sum of the OP

$$\begin{align}s_{\log (2),1}=\log(2)=\sum _{n=1}^{\infty } \frac{2 n+1}{2^{2n}} \left[\frac{n-1}{2^{2n}} -(n-2)\right]\zeta (2 n+1)\end{align}\tag{b1} $$

actually $s_{\log (2),1}=2 \log (4)-\log (8)$

$$\begin{align} s_{\log (2),2}=\log(2)=\frac{1}{5}\sum _{n=1}^{\infty }\frac{2 n+1}{2^{2 n}} \left[\frac{n-1}{2^{2 n}}-(n-4)\right] \zeta (2 n+1)\end{align}\tag{b2} $$

actually $s_{\log (2),2}=\frac{1}{5} (4 \log (4)-\log (8))$

$$\begin{align} s_{\zeta(3)}=\zeta(3)=\frac{1}{105}\sum _{n=1}^{\infty }\frac{2 n+1}{2^{2n-1}} \left[\frac{22(n-1)}{2^{2n}} +8 n + 33 \right] \zeta (2 n+1) \end{align}\tag{c}$$

actually, $s_{\zeta(3)} = \frac{1}{105} (105 \zeta (3)+66 \log (4)-44 \log (8))$

It came as a surpise to me that, with the indicated method of linear combination, I obtained a different representations of $\log(2)$ than that of the OP. I found, however, that the sums provided here all result in more complicated expressions (than necessary) which simplify due the properties of $\log$.

§3. A simpler filter comes from the relations

$$\begin{align} s_1 &:=\sum _{n=1}^{\infty } \frac{2 n+1}{2^{2 n}} \zeta (2 n+1)=\log(4)\\ s_2 & := \sum _{n=1}^{\infty } \frac{2 n+1}{2^{4 n}} \zeta (2 n+1) = \log (8)-2 C\\ s_3 &:=\sum _{n=1}^{\infty } \frac{(2 n+1) n \zeta (2 n+1)}{2^{2 n}}=\frac{7}{4} \zeta(3) \end {align}\tag{d1}$$

whence

$$\log(2) = \frac{1}{2} s_1, C = \frac{1}{4}(3s_1 - 2 s_2), \zeta(3) = \frac{4}{7} s_3\tag{d2}$$

Notice that we obtain here a simpler formula for Catalan's number than in $(a)$

$$C=\sum _{n=1}^{\infty } \frac{2 n+1}{2^{2 n+2}} \left(3-\frac{2}{2^{2 n}}\right) \zeta (2 n+1)\tag{d3}$$

since here is no factor $n$ in the summand.

As $\zeta(3)$ apears also on the r.h.s of the last equation od $(d2)$ we can eliminate it and find

$$\zeta(3) = \sum _{n=2}^{\infty } \frac{2 n+1}{2^{2 n}} n \zeta (2 n+1)\tag{e1}$$

which expresses $\zeta(3)$ as a sum over higher $\zeta$ functions.

Similarly we can find

$$\zeta(5) = \frac{8}{357} \sum _{n=3}^{\infty } \frac{2 n+1}{2^{2 n}} \left(8 n^3-29 n\right) \zeta (2 n+1)\tag{e2}$$

and so on.

§4. Higher derivatives, leading to higher powers of $k$ in the generating sum, are expressible in terms of the generalized harmonic number $H_{z}^{(m)}$ for the $m$-th derivative.

For $z=1/2$ or $z=1/4$ these are expressible by the Hurwitz zeta function. This describes the set of closed results for the sums of the type $(1)$.

§5. A further generalization would be a different choice of the values of $z$.