Conjectured closed forms for Eisenstein-like series

Question

This question is related to: Eisenstein sum. Being $q=e^{\pi}$, we have also:

\begin{align*} \sum_{n=1}^{\infty}\frac{n(q^{n}(-1)^{n}+1)}{q^{2n}+2(-1)^{n}q^{n}+1}=-\frac{1}{24}\tag{1}, \end{align*} \begin{align*} \sum_{n=1}^{\infty}\frac{n(q^{n/2}\cos{(\pi n/2)}+1)}{q^{n}+2\cos{(\pi n /2)}q^{n/2}+1}=-\frac{1}{24}.\tag{2} \end{align*} And a lot more of this kind can be obtained. I have no proofs for (1) and (2). Context: If $q=e^{\pi}$ it is known that: \begin{align*} \sum_{n=1}^{\infty}\frac{2n-1}{q^{2n-1}+1}=\sum_{2\nmid n}\frac{n}{q^n+1}=\frac{1}{24}.\tag{3} \end{align*} If $Q=-e^{-\pi}$ we have: \begin{align*} \sum_{2\nmid n}\frac{n}{q^n+1}=-\sum_{2\nmid n}\frac{nQ^n}{1-Q^n}=-\sum_{2\nmid n}\sum_{k=1}^{\infty}nQ^{nk} \\=-\sum_{l=1}^{\infty}\sigma_{1}^{odd}(l)Q^{l}=\sum_{l=1}^{\infty}\left(2\sigma_{1}(l)-\sigma{(2l)} \right)Q^{l}, \end{align*} where $\sigma_{1}$ is the sum divisors function and we have use $\sigma_{1}^{odd}(m)=\sigma_{1}(2m)-2\sigma_{1}(m)$. We know from Poisson's summation formula that: \begin{align*} f(\tau)=\sum_{l=1}^{\infty}\sigma_{1}(l)e^{2\pi i l \tau}\hspace{.3cm}\text{ satisfies } \hspace{.3cm}f(-1/\tau)=\tau^2f(\tau)+\frac{1}{24}-\frac{\tau^2}{24}+\frac{i \tau}{4\pi}.\tag{4} \end{align*} For each $\tau$ in the upper complex semiplane. With this notation $(3)$ is equivalent to: \begin{align*} \sum_{2\nmid n}\frac{n}{q^n+1}=2f(\frac{1+i}{2})-\frac{1}{2}f(\frac{1+i}{4})-\frac{1}{2}f(\frac{i-1}{4}).\tag{5} \end{align*} In this case we don't need the link to theta functions. First taking $\tau=i$ in $(4)$ we have: \begin{align*} f(i)=\frac{1}{24}-\frac{1}{8\pi}.\tag{6} \end{align*} Now setting $\tau=i-1$: \begin{align*} f(\frac{1+i}{2})=f(-\frac{1}{i-1}) \\=(i-1)^2f(i)+\frac{1}{24}-\frac{(i-1)^2}{24}+\frac{i(i-1)}{4\pi} \\=\frac{1}{24}-\frac{1}{4\pi}.\tag{7} \end{align*} For the remaining, setting $\tau=2i\pm2$: \begin{align*} f(\frac{i\mp 1}{4})=\pm 8if(2i)+\frac{1}{24}\mp \frac{i}{3}+\frac{i(2i\pm2)}{4\pi}. \end{align*} And making the sum for boths signs in this last equality: \begin{align*} f(\frac{i+1}{4})+f(\frac{i-1}{4})=\frac{1}{12}-\frac{1}{\pi}.\tag{8} \end{align*} Finally setting $(7)$ and $(8)$ in $(5)$ we get $\frac{1}{24}$.

Question: Can we prove these sums using only the functional equations associated to each one and avoiding the use of theta functions?

Updated: Since Paramanand has solved both sums. Here is the general conjecture:

\begin{align*} \sum_{n=1}^{\infty}\frac{n(e^{\frac{\pi n}{z}}\cos{(\frac{\pi n}{z})}+1)}{e^{\frac{2\pi n}{z}}+2e^{\frac{\pi n}{z}}\cos{(\frac{\pi n}{z})}+1}=-\frac{1}{24},\tag{2.1} \end{align*} when $z \in \mathbb{N}$. For example for $z=10$: \begin{align*} \sum_{n=1}^{\infty}\frac{n(e^{\frac{\pi n}{10}}\cos{(\frac{\pi n}{10})}+1)}{e^{\frac{\pi n}{5}}+2e^{\frac{\pi n}{10}}\cos{(\frac{\pi n}{10})}+1}=-\frac{1}{24}.\tag{2.2} \end{align*} The function is not constant but it has an intringued behaviour.

Answer 1

The sums in question can be evaluated in closed form using the functional equations for Eisenstein series $E_2$ defined by $$E_2(\tau) =1-24\sum_{n\geq 1}\frac{nq^n}{1-q^n}\tag{1}$$ where $q=\exp(2\pi i\tau) $ and $\tau$ is a complex number with positive imaginary part.

The functional equations for $E_2$ are $$E_2(\tau+1)=E_2(\tau)\tag{2}$$ and $$E_2(-1/\tau)=\tau^2 E_2(\tau)-\frac{6i\tau}{\pi}\tag{3}$$ Putting $\tau=i$ in $(3)$ we get $E_2(i)=\frac{3}{\pi}$.

Next we put $\tau=(i-1)/2$ in $(3)$ and note that $$E_2(-1/\tau)=E_2(1+i)=E_2(i)=\frac{3}{\pi}$$ and hence $$-\frac{i} {2}\cdot E_2((i-1)/2)-\frac{3i(i-1)}{\pi}=\frac{3}{\pi}$$ or $$E_2((i-1)/2)=\frac{6}{\pi}\tag{4}$$ Next we can note that the first sum in question can be written as $$S=\sum_{n\geq 1}\frac{nq^n}{1+q^n}\tag{5}$$ where $q=-e^{-\pi} =\exp(2\pi i\tau) $ with $\tau=(i-1)/2$. A little algebra shows that $$S=\sum_{n\geq 1}\frac{nq^n}{1-q^n}-2\sum_{n\geq 1}\frac{nq^{2n}}{1-q^{2n}}$$ which further equals $$\frac{1-E_2(\tau)}{24}-\frac{1-E_2(2\tau)}{12}=-\frac{1}{24}+\frac{2E_2(2\tau)-E_2(\tau)}{24}$$ We have $E_2(\tau)=6/\pi$ and $E_2(2\tau)=E_2(i-1)=E_2(i)=3/\pi$ so that the desired sum is $-1/24$.

For second sum we need to note that $\cos(n\pi/2)$ takes values $0,-1,0,1$ repeatedly as $n$ takes values $1,2,3,4,\dots$. Hence the sum equals $$\sum_{n\geq 1}\frac{nq^n}{1+q^n}+\sum_{n\geq 1}\frac{2nq^{2n}}{1+q^{2n}}-2\sum_{n\geq 1}\frac{(2n-1)q^{2n-1}}{1-q^{2n-1}}$$ which further simplifies to $$\sum_{n\geq 1}\frac{2nq^{2n}}{1+q^{2n}}-\sum_{n\geq 1}\frac{(2n-1)q^{2n-1}}{1-q^{2n-1}}$$ where $q=e^{-\pi} $.

This can be expressed in terms of $E_2$ as $$-\frac{1}{24}+\frac{4E_2(4\tau)-4E_2(2\tau)+E_2(\tau)}{24}$$ where $\tau=i/2$. We already know that $E_2(2\tau)=E_2(i)=3/\pi$ and putting $\tau=i/2$ in $(3) $ we get $$4E_2(4\tau)+E_2(\tau)=\frac{12}{\pi}$$ so that the second sum is also $-1/24$.

---

The following deals with the generalisation at the end of the question.

Let $\tau$ be a complex number with positive imaginary part and let $x$ be real and let $q=\exp(2\pi i\tau) $ so that $|q|<1$ and consider the product $$f(q, x)=\prod_{n=1}^{\infty}(1+2q^n\cos(2\pi nx) +q^{2n})\tag{6}$$ Using logarithmic differentiation with respect to $q$ we get $$\frac {q} {2}\frac{d}{dq}\log f(q,x)=\sum_{n=1}^{\infty}\frac{n(q^n\cos(2\pi nx) +q^{2n})}{1+2q^n\cos(2\pi nx) +q^{2n}} \tag{7}$$ The generalised sum in question is the value of above series with $q=e^{-\pi/z}, x=1/2z$ so that $\tau=i/2z$.

Let $u=\tau+x, v=\tau-x$ and then we can write $$f(q, x) =\prod_{n=1}^{\infty} (1+\exp(2\pi i nu)) (1+\exp(2\pi i nv)) $$ which can be written in terms of Dedekind's eta function as $$f(q, x) =\exp(-\pi i\tau/6)\frac{\eta(2u)\eta(2v)}{\eta(u)\eta(v)}\tag{8}$$ where $$\eta(\tau) =\exp(\pi i\tau/12)\prod_{n=1}^{\infty}(1-\exp(2\pi in \tau)) \tag{9}$$ We have by definition $$E_2(\tau)=\frac{12}{\pi i} \frac{d} {d\tau} \log\eta(\tau) \tag{10}$$ Replacing $\tau$ with $2\tau$ we get $$E_2(2\tau)=\frac{6}{\pi i} \frac{d} {d\tau} \log\eta(2\tau)\tag{11}$$ Note that replacing $\tau$ with $\tau+k$ (where $k$ is constant) in above formulas does not change the differentiation operator $d/d\tau$ so these formulas work if we write $u, v$ for $\tau$ and keep the differentiation operator as $d/d\tau$.

Next we also need to note that $$q\frac{d} {dq} =\frac{1}{2\pi i} \frac{d} {d\tau} $$ and hence the sum, say $g(q,x) $, in $(7)$ can be written as $$\frac{1}{4\pi i} \frac{d} {d\tau} \log f(q, x) $$ and using $(8)$ we get $$g(q, x) =-\frac{1}{24}+\frac{2E_2(2u)+2E_2(2v)-E_2(u)-E_2(v)}{48}\tag{7}$$ using $(10),(11)$. We have $$u=\tau+x=\frac{i+1}{2z},v=\frac{i-1}{2z}$$ For a positive integer $z$ we can observe via $(3)$ that $$E_2(u)=E_2((i+1)/2z)=(z(i-1))^2E_2(z(i-1))-\frac{6iz(i-1)}{\pi}$$ and $$E_2(v)=(z(i+1))^2E_2(z(i+1))-\frac{6iz(i+1)}{\pi}$$ so that $$E_2(u)+E_2(v)=\frac{12z}{\pi}$$ We further need to note that $$E_2(2u)=E_2((1+i)/z)=(z(i-1)/2)^2E_2(z(i-1)/2)-\frac{3iz(i-1)}{\pi}$$ and $$E_2(2v)=(z(i+1)/2)^2E_2(z(i+1)/2)-\frac{3iz(i+1)}{\pi}$$ and then $$E_2(2u)+E_2(2v)=\frac{6z}{\pi}$$ so the sum is indeed $-1/24$.

Since $E_2$ is periodic with period $1$ (see $(2)$) its value remains same when the argument is shifted by an integer and in the above derivation this fact has been used taking into account that $z$ is an integer. The derivation and the result don't hold if $z $ is not a positive integer.