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"Can one modify the known appraoches to classical Ramanujan-type series for $1/\pi$ to prove the above general conjecture?" -- Yes. Actually much more can be produced.

The following notations are classical (see for example, Pi and AGM by Browein). For $0\leq s < 1/2$, let $$K_s(k) = \frac{\pi}{2} {_2F_1}(\frac{1}{2}-s,\frac{1}{2}+s;1;k^2) \qquad E_s(k) = \frac{\pi}{2} {_2F_1}(-\frac{1}{2}-s,\frac{1}{2}+s;1;k^2)$$ $k' = \sqrt{1-k^2}$, $K_s'(k) = K_s(k')$. $$\alpha_s(r) = \frac{\pi}{4K_s^2} \frac{\cos \pi s}{1+2s} - \sqrt{r}(\frac{E_s}{K_s}-1)$$ for positive rational $N$, let $k_{s,N}$ be such that $K_s'(k_{s,N}) = \sqrt{N}K_s(k_{s,N})$. It can be shown $$\tag{1}\frac{\cos \pi s}{\pi (1+2s)} = \frac{\sqrt{N}k_N k_N'^2}{1+2s} \frac{4K_s}{\pi^2} \frac{d K_s}{dk} + [\alpha_s(N)-\sqrt{N}k_N^2] \frac{4K_s^2}{\pi^2}$$ $$\tag{2}0 = \frac{\sqrt{N}k_N k_N'^2}{2(1+2s)} \frac{d}{dk} (K_s K'_s) + [\alpha_s(N)-\sqrt{N}k_N^2] K_s K'_s$$

When $s\in \{0,1/3,1/6,1/4\}$, $N$ positive rational, all quantities in $(1), (2)$ except $K, K'$ are algebraic number: because they are values of modular (or derivatives thereof) at CM points. These are of course, well-known conclusions.

For the rest, assuming $0\leq s < 1/2$ is enough, for $0\leq k < \frac{1}{\sqrt{2}}$, we have

$$\tag{*}\begin{align*}K_s(k)^2 &= \frac{\pi^2}{4}\sum_{n\geq 0} c_n (2kk')^{2n} \\ K_s(k) K'_s(k) &= \frac{\pi \cos s\pi}{4}\sum_{n\geq 0} c_n (d_n - 2\log(2kk')) (2kk')^{2n} \\ K'_s(k)^2 + K_s(k)^2&= \frac{\cos^2(\pi s)}{4} \sum_{n\geq 0} c_n \left(-4 d_n \log (2kk')+e_n+4\log^2(2kk')\right) (2kk')^{2n} \end{align*}$$ here, in terms of Pochhammer symbol and polygamma function, $$\begin{aligned} c_n &= \frac{(1/2-s)_n (1/2+s)_n (1/2)_n}{(n!)^3} \\ d_n &= -\psi(n-s+\frac{1}{2})-\psi(n+s+\frac{1}{2})+3 \psi(n+1)-\psi(n+\frac{1}{2})\end{aligned}$$ A uniform explanation of all three formulas $(*)$: they are solutions of a 3rd order ODE, whose indicial equation at origin has triple root. The $\log, \log^2$ comes from general theory of Frobenius method.

Inserting the first series into $(1)$, gives our familiar $1/\pi$-formula: $$\frac{\cos \pi s}{\pi (1+2s)} = \sum_{n\geq 0} c_n (A_{s,N} n + B_{s,N}) (2kk')^{2n}$$ with $$A_{s,N} = \frac{\sqrt{N}}{1+2s} (-k^2+k'^2) \qquad B_s(N) = \alpha_s(N) - \sqrt{N}k_N^2$$

If one uses instead the expansion of $K_s K'_s$, and plug it into $(2)$: $$ 0 = -2\log(2kk') \frac{\cos \pi s}{\pi(1+2s)} + \sum_{n\geq 0} c_n (2kk')^{2n} (-A_{s,N} + (n A_{s,N} + B_{s,N}) d_n)$$

When $s=0$, this reduces to conjecture $(I)$ of OP: $d_n =3\log 4 -6 \left(H_{2 n}-H_n\right)$; conjectures $(II), (III), (IV)$ are from $s=1/6, 1/4, 1/3$ respectively. Q.E.D.

If one uses the expansion for $K'_s(k)^2 + K_s(k)^2$, one obtained similar series involving second order harmonic numbers.

"Can one modify the known appraoches to classical Ramanujan-type series for $1/\pi$ to prove the above general conjecture?" -- Yes. 

Actually much more can be proved.

We use classical notations (see for example, Pi and AGM by Browein, Chap. 5.5). For $0\leq s < 1/2$, let $$K_s(k) = \frac{\pi}{2} {_2F_1}(\frac{1}{2}-s,\frac{1}{2}+s;1;k^2) \qquad E_s(k) = \frac{\pi}{2} {_2F_1}(-\frac{1}{2}-s,\frac{1}{2}+s;1;k^2)$$ $k' = \sqrt{1-k^2}$, $K_s'(k) = K_s(k')$. $$\alpha_s(r) = \frac{\pi}{4K_s^2} \frac{\cos \pi s}{1+2s} - \sqrt{r}(\frac{E_s}{K_s}-1)$$ for positive rational $N$, let $k_{N}$ be such that $K_s'(k_{N}) = \sqrt{N}K_s(k_{N})$. It can be shown $$\tag{1}\frac{\cos \pi s}{\pi (1+2s)} = \frac{\sqrt{N}k_N k_N'^2}{1+2s} \frac{4K_s}{\pi^2} \frac{d K_s}{dk} + [\alpha_s(N)-\sqrt{N}k_N^2] \frac{4K_s^2}{\pi^2}$$ $$\tag{2}0 = \frac{\sqrt{N}k_N k_N'^2}{2(1+2s)} \frac{d}{dk} (K_s K'_s) + [\alpha_s(N)-\sqrt{N}k_N^2] K_s K'_s$$

When $s\in \{0,1/3,1/6,1/4\}$, $N$ positive rational, all quantities in $(1), (2)$ except $K, K'$, are algebraic number: because they are values of modular (or derivatives thereof) functions at CM points. These are of course, well-known results.

For the rest, assuming $0\leq s < 1/2$ is enough. If $0\leq k < \frac{1}{\sqrt{2}}$, we have

$$\tag{*}\begin{align*}K_s(k)^2 &= \frac{\pi^2}{4}\sum_{n\geq 0} c_n (2kk')^{2n} \\ K_s(k) K'_s(k) &= \frac{\pi \cos s\pi}{4}\sum_{n\geq 0} c_n (d_n - 2\log(2kk')) (2kk')^{2n} \\ K'_s(k)^2 + K_s(k)^2&= \frac{\cos^2(\pi s)}{4} \sum_{n\geq 0} c_n \left(-4 d_n \log (2kk')+e_n+4\log^2(2kk')\right) (2kk')^{2n} \end{align*}$$ here, in terms of Pochhammer symbol and polygamma function, $$\begin{aligned} c_n &= \frac{(1/2-s)_n (1/2+s)_n (1/2)_n}{(n!)^3} \\ d_n &= -\psi(n-s+\frac{1}{2})-\psi(n+s+\frac{1}{2})+3 \psi(n+1)-\psi(n+\frac{1}{2})\\ e_n &= \text{ certain expression in digamma and trigamma}\end{aligned}$$ An intuitive explanation of all formulas $(*)$: LHS are solutions of a 3rd order ODE, whose indicial equation at origin has triple root. The $\log, \log^2$ comes from general theory of Frobenius method, and polygamma are typical in the expansion of such solutions.

Inserting the first series into $(1)$, gives our familiar $1/\pi$-formula: $$\frac{\cos \pi s}{\pi (1+2s)} = \sum_{n\geq 0} c_n (A_{s,N} n + B_{s,N}) (2kk')^{2n}$$ with $A_{s,N} = \frac{\sqrt{N}}{1+2s} (-k^2+k'^2)$ and $B_s(N) = \alpha_s(N) - \sqrt{N}k_N^2$.

If one uses instead the expansion of $K_s K'_s$, and plug it into $(2)$: $$ 0 = -2\log(2kk') \frac{\cos \pi s}{\pi(1+2s)} + \sum_{n\geq 0} c_n (2kk')^{2n} (-A_{s,N} + (n A_{s,N} + B_{s,N}) d_n)$$

When $s=0$, this reduces to conjecture (I) of OP: $d_n =3\log 4 -6 \left(H_{2 n}-H_n\right)$, so above expression becomes $$\frac{1}{\pi} \log [4^3(2kk')^2] = \sum_{n\geq 0} c_n (2kk')^{2n} [A_{0,N} + 6(H_{2n}-H_n) (nA_{0,N} + B_{0,N})]$$ matching the form conjectured by OP.

Conjectures (II), (III), (IV) follow from $s=1/6, 1/4, 1/3$ respectively. Q.E.D.

If one uses the expansion for $K'_s(k)^2 + K_s(k)^2$, one obtains similar series involving second order harmonic numbers.

"Can one modify the known appraoches to classical Ramanujan-type series for $1/\pi$ to prove the above general conjecture?" -- Yes. Actually much more can be produced.

The following notations are classical (see for example, Pi and AGM by Browein). For $0\leq s < 1/2$, let $$K_s(k) = \frac{\pi}{2} {_2F_1}(\frac{1}{2}-s,\frac{1}{2}+s;1;k^2) \qquad E_s(k) = \frac{\pi}{2} {_2F_1}(-\frac{1}{2}-s,\frac{1}{2}+s;1;k^2)$$ $k' = \sqrt{1-k^2}$, $K_s'(k) = K_s(k')$. $$\alpha_s(r) = \frac{\pi}{4K_s^2} \frac{\cos \pi s}{1+2s} - \sqrt{r}(\frac{E_s}{K_s}-1)$$ for positive rational $N$, let $k_{s,N}$ be such that $K_s'(k_{s,N}) = \sqrt{N}K_s(k_{s,N})$. Then, from Legendre's relation $$E_s K_s' + K_s E_s' - K_s K_s' = \frac{\pi}{2} \frac{\cos \pi s}{1+2s}$$ we have $$\tag{1}\frac{\cos \pi s}{\pi (1+2s)} = \frac{\sqrt{N}k_N k_N'^2}{1+2s} \frac{4K_s}{\pi^2} \frac{d K_s}{dk} + [\alpha_s(N)-\sqrt{N}k_N^2] \frac{4K_s^2}{\pi^2}$$ $$\tag{2}0 = \frac{\sqrt{N}k_N k_N'^2}{2(1+2s)} \frac{d}{dk} (K_s K'_s) + [\alpha_s(N)-\sqrt{N}k_N^2] K_s K'_s$$

When $s\in \{0,1/3,1/6,1/4\}$, $N$ positive rational, all quantities in $(1), (2)$ except $K, K'$ are algebraic number: because they are values of modular (or derivatives thereof) at CM points. These are of course, well-known conclusions.

For the rest, assuming $0\leq s < 1/2$ is enough, for $0\leq k < \frac{1}{\sqrt{2}}$, we have

$$\tag{*}\begin{align*}K_s(k)^2 &= \frac{\pi^2}{4}\sum_{n\geq 0} c_n (2kk')^{2n} \\ K_s(k) K'_s(k) &= \frac{\pi \cos s\pi}{4}\sum_{n\geq 0} c_n (d_n - 2\log(2kk')) (2kk')^{2n} \\ K'_s(k)^2 + K_s(k)^2&= \frac{\cos^2(\pi s)}{4} \sum_{n\geq 0} c_n \left(-4 d_n \log (2kk')+e_n+4\log^2(2kk')\right) (2kk')^{2n} \end{align*}$$ here, in terms of Pochhammer symbol and polygamma function, $$\begin{aligned} c_n &= \frac{(1/2-s)_n (1/2+s)_n (1/2)_n}{(n!)^3} \\ d_n &= -\psi ^{(0)}(n-s+\frac{1}{2})-\psi ^{(0)}(n+s+\frac{1}{2})+3 \psi ^{(0)}(n+1)-\psi ^{(0)}(n+\frac{1}{2})\end{aligned}$$ A uniform explanation of all three formulas $(*)$: they are solutions of a 3rd order ODE, whose indicial equation at origin has triple root. The $\log, \log^2$ comes from general theory of Frobenius method.

Inserting the first series into $(1)$, gives our familiar $1/\pi$-formula: $$\frac{\cos \pi s}{\pi (1+2s)} = \sum_{n\geq 0} c_n (A_{s,N} n + B_{s,N}) (2kk')^{2n}$$ with $$A_{s,N} = \frac{\sqrt{N}}{1+2s} (-k^2+k'^2) \qquad B_s(N) = \alpha_s(N) - \sqrt{N}k_N^2$$

If one uses instead the second series, and plug it into $(2)$: $$ 0 = -2\log(2kk') \frac{\cos \pi s}{\pi(1+2s)} + \sum_{n\geq 0} c_n (2kk')^{2n} (-A_{s,N} + (n A_{s,N} + B_{s,N}) d_n)$$

When $s=0$, this reduces to conjecture $(I)$ of OP: $d_n$ becomes $3\log 4 -6 \left(H_{2 n}-H_n\right)$; conjectures $(II), (III), (IV)$ comes from $s=1/6, 1/4, 1/3$ respectively. Q.E.D.

If one uses the expansion for $K'_s(k)^2 + K_s(k)^2$, one obtained similar series involving second order harmonic numbers.

"Can one modify the known appraoches to classical Ramanujan-type series for $1/\pi$ to prove the above general conjecture?" -- Yes. Actually much more can be produced.

The following notations are classical (see for example, Pi and AGM by Browein). For $0\leq s < 1/2$, let $$K_s(k) = \frac{\pi}{2} {_2F_1}(\frac{1}{2}-s,\frac{1}{2}+s;1;k^2) \qquad E_s(k) = \frac{\pi}{2} {_2F_1}(-\frac{1}{2}-s,\frac{1}{2}+s;1;k^2)$$ $k' = \sqrt{1-k^2}$, $K_s'(k) = K_s(k')$. $$\alpha_s(r) = \frac{\pi}{4K_s^2} \frac{\cos \pi s}{1+2s} - \sqrt{r}(\frac{E_s}{K_s}-1)$$ for positive rational $N$, let $k_{s,N}$ be such that $K_s'(k_{s,N}) = \sqrt{N}K_s(k_{s,N})$. It can be shown $$\tag{1}\frac{\cos \pi s}{\pi (1+2s)} = \frac{\sqrt{N}k_N k_N'^2}{1+2s} \frac{4K_s}{\pi^2} \frac{d K_s}{dk} + [\alpha_s(N)-\sqrt{N}k_N^2] \frac{4K_s^2}{\pi^2}$$ $$\tag{2}0 = \frac{\sqrt{N}k_N k_N'^2}{2(1+2s)} \frac{d}{dk} (K_s K'_s) + [\alpha_s(N)-\sqrt{N}k_N^2] K_s K'_s$$

When $s\in \{0,1/3,1/6,1/4\}$, $N$ positive rational, all quantities in $(1), (2)$ except $K, K'$ are algebraic number: because they are values of modular (or derivatives thereof) at CM points. These are of course, well-known conclusions.

For the rest, assuming $0\leq s < 1/2$ is enough, for $0\leq k < \frac{1}{\sqrt{2}}$, we have

$$\tag{*}\begin{align*}K_s(k)^2 &= \frac{\pi^2}{4}\sum_{n\geq 0} c_n (2kk')^{2n} \\ K_s(k) K'_s(k) &= \frac{\pi \cos s\pi}{4}\sum_{n\geq 0} c_n (d_n - 2\log(2kk')) (2kk')^{2n} \\ K'_s(k)^2 + K_s(k)^2&= \frac{\cos^2(\pi s)}{4} \sum_{n\geq 0} c_n \left(-4 d_n \log (2kk')+e_n+4\log^2(2kk')\right) (2kk')^{2n} \end{align*}$$ here, in terms of Pochhammer symbol and polygamma function, $$\begin{aligned} c_n &= \frac{(1/2-s)_n (1/2+s)_n (1/2)_n}{(n!)^3} \\ d_n &= -\psi(n-s+\frac{1}{2})-\psi(n+s+\frac{1}{2})+3 \psi(n+1)-\psi(n+\frac{1}{2})\end{aligned}$$ A uniform explanation of all three formulas $(*)$: they are solutions of a 3rd order ODE, whose indicial equation at origin has triple root. The $\log, \log^2$ comes from general theory of Frobenius method.

Inserting the first series into $(1)$, gives our familiar $1/\pi$-formula: $$\frac{\cos \pi s}{\pi (1+2s)} = \sum_{n\geq 0} c_n (A_{s,N} n + B_{s,N}) (2kk')^{2n}$$ with $$A_{s,N} = \frac{\sqrt{N}}{1+2s} (-k^2+k'^2) \qquad B_s(N) = \alpha_s(N) - \sqrt{N}k_N^2$$

If one uses instead the expansion of $K_s K'_s$, and plug it into $(2)$: $$ 0 = -2\log(2kk') \frac{\cos \pi s}{\pi(1+2s)} + \sum_{n\geq 0} c_n (2kk')^{2n} (-A_{s,N} + (n A_{s,N} + B_{s,N}) d_n)$$

When $s=0$, this reduces to conjecture $(I)$ of OP: $d_n =3\log 4 -6 \left(H_{2 n}-H_n\right)$; conjectures $(II), (III), (IV)$ are from $s=1/6, 1/4, 1/3$ respectively. Q.E.D.

If one uses the expansion for $K'_s(k)^2 + K_s(k)^2$, one obtained similar series involving second order harmonic numbers.

"Can one modify the known appraoches to classical Ramanujan-type series for 1/π to prove the above general conjecture?" -- Yes.

The following notations are classical (see for example, Pi and AGM by Browein). For $0\leq s < 1/2$, let $$K_s(k) = \frac{\pi}{2} {_2F_1}(\frac{1}{2}-s,\frac{1}{2}+s;1;k^2) \qquad E_s(k) = \frac{\pi}{2} {_2F_1}(-\frac{1}{2}-s,\frac{1}{2}+s;1;k^2)$$ $k' = \sqrt{1-k^2}$, $K_s'(k) = K_s(k')$. $$\alpha_s(r) = \frac{\pi}{4K_s^2} \frac{\cos \pi s}{1+2s} - \sqrt{r}(\frac{E_s}{K_s}-1)$$ for positive rational $N$, let $k_{s,N}$ be such that $K_s'(k_{s,N}) = \sqrt{N}K_s(k_{s,N})$. Then, from Legendre's relation $$E_s K_s' + K_s E_s' - K_s K_s' = \frac{\pi}{2} \frac{\cos \pi s}{1+2s}$$ we have $$\tag{1}\frac{\cos \pi s}{\pi (1+2s)} = \frac{\sqrt{N}k_N k_N'^2}{1+2s} \frac{4K_s}{\pi^2} \frac{d K_s}{dk} + [\alpha_s(N)-\sqrt{N}k_N^2] \frac{4K_s^2}{\pi^2}$$ $$\tag{2}0 = \frac{\sqrt{N}k_N k_N'^2}{2(1+2s)} \frac{d}{dk} (K_s K'_s) + [\alpha_s(N)-\sqrt{N}k_N^2] K_s K'_s$$

When $s\in \{0,1/3,1/6,1/4\}$, $N$ positive rational, all quantities in $(1), (2)$ except $K, K'$ are algebraic number: because they are values of modular (or derivatives thereof) at CM points. These are of course, well-known conclusions.

For the rest, assuming $0\leq s < 1/2$ is enough, for $0\leq k < \frac{1}{\sqrt{2}}$,

$$\begin{aligned}K_s(k)^2 &= \frac{\pi^2}{4}\sum_{n\geq 0} c_n (2kk')^{2n} \\ K_s(k) K'_s(k) &= \frac{\pi \cos s\pi}{4}\sum_{n\geq 0} c_n (d_n - 2\log(2kk')) (2kk')^{2n} \\ K'_s(k)^2 + K_s(k)^2&= \frac{\cos^2(\pi s)}{4} \sum_{n\geq 0} c_n \left(-4 d_n \log (2kk')+e_n+4\log^2(2kk')\right) (2kk')^{2n} \end{aligned}$$ here, in terms of Pochhammer symbol and polygamma function, $$\begin{aligned} c_n &= \frac{(1/2-s)_n (1/2+s)_n (1/2)_n}{(n!)^3} \\ d_n &= -\psi ^{(0)}(n-s+\frac{1}{2})-\psi ^{(0)}(n+s+\frac{1}{2})+3 \psi ^{(0)}(n+1)-\psi ^{(0)}(n+\frac{1}{2})\end{aligned}$$

Inserting the first series into $(1)$, gives our familiar $1/\pi$-formula: $$\frac{\cos \pi s}{\pi (1+2s)} = \sum_{n\geq 0} c_n (A_{s,N} n + B_{s,N}) (2kk')^{2n}$$ with $$A_{s,N} = \frac{\sqrt{N}}{1+2s} (-k^2+k'^2) \qquad B_s(N) = \alpha_s(N) - \sqrt{N}k_N^2$$

If one uses instead the second series, and plug it into $(2)$: $$ 0 = -2\log(2kk') \frac{\cos \pi s}{\pi(1+2s)} + \sum_{n\geq 0} c_n (2kk')^{2n} (-A_{s,N} + (n A_{s,N} + B_{s,N}) d_n)$$

When $s=0$, this reduces to conjecture $(I)$ of OP: $d_n$ becomes $3\log 4 -6 \left(H_{2 n}-H_n\right)$; conjectures $(II), (III), (IV)$ comes from $s=1/6, 1/4, 1/3$ respectively. Q.E.D.

If one uses the expansion for $K'_s(k)^2 + K_s(k)^2$, one obtained similar series involving second order harmonic numbers.

"Can one modify the known appraoches to classical Ramanujan-type series for $1/\pi$ to prove the above general conjecture?" -- Yes. Actually much more can be produced.

The following notations are classical (see for example, Pi and AGM by Browein). For $0\leq s < 1/2$, let $$K_s(k) = \frac{\pi}{2} {_2F_1}(\frac{1}{2}-s,\frac{1}{2}+s;1;k^2) \qquad E_s(k) = \frac{\pi}{2} {_2F_1}(-\frac{1}{2}-s,\frac{1}{2}+s;1;k^2)$$ $k' = \sqrt{1-k^2}$, $K_s'(k) = K_s(k')$. $$\alpha_s(r) = \frac{\pi}{4K_s^2} \frac{\cos \pi s}{1+2s} - \sqrt{r}(\frac{E_s}{K_s}-1)$$ for positive rational $N$, let $k_{s,N}$ be such that $K_s'(k_{s,N}) = \sqrt{N}K_s(k_{s,N})$. Then, from Legendre's relation $$E_s K_s' + K_s E_s' - K_s K_s' = \frac{\pi}{2} \frac{\cos \pi s}{1+2s}$$ we have $$\tag{1}\frac{\cos \pi s}{\pi (1+2s)} = \frac{\sqrt{N}k_N k_N'^2}{1+2s} \frac{4K_s}{\pi^2} \frac{d K_s}{dk} + [\alpha_s(N)-\sqrt{N}k_N^2] \frac{4K_s^2}{\pi^2}$$ $$\tag{2}0 = \frac{\sqrt{N}k_N k_N'^2}{2(1+2s)} \frac{d}{dk} (K_s K'_s) + [\alpha_s(N)-\sqrt{N}k_N^2] K_s K'_s$$

When $s\in \{0,1/3,1/6,1/4\}$, $N$ positive rational, all quantities in $(1), (2)$ except $K, K'$ are algebraic number: because they are values of modular (or derivatives thereof) at CM points. These are of course, well-known conclusions.

For the rest, assuming $0\leq s < 1/2$ is enough, for $0\leq k < \frac{1}{\sqrt{2}}$, we have

$$\tag{*}\begin{align*}K_s(k)^2 &= \frac{\pi^2}{4}\sum_{n\geq 0} c_n (2kk')^{2n} \\ K_s(k) K'_s(k) &= \frac{\pi \cos s\pi}{4}\sum_{n\geq 0} c_n (d_n - 2\log(2kk')) (2kk')^{2n} \\ K'_s(k)^2 + K_s(k)^2&= \frac{\cos^2(\pi s)}{4} \sum_{n\geq 0} c_n \left(-4 d_n \log (2kk')+e_n+4\log^2(2kk')\right) (2kk')^{2n} \end{align*}$$ here, in terms of Pochhammer symbol and polygamma function, $$\begin{aligned} c_n &= \frac{(1/2-s)_n (1/2+s)_n (1/2)_n}{(n!)^3} \\ d_n &= -\psi ^{(0)}(n-s+\frac{1}{2})-\psi ^{(0)}(n+s+\frac{1}{2})+3 \psi ^{(0)}(n+1)-\psi ^{(0)}(n+\frac{1}{2})\end{aligned}$$ A uniform explanation of all three formulas $(*)$: they are solutions of a 3rd order ODE, whose indicial equation at origin has triple root. The $\log, \log^2$ comes from general theory of Frobenius method.

Inserting the first series into $(1)$, gives our familiar $1/\pi$-formula: $$\frac{\cos \pi s}{\pi (1+2s)} = \sum_{n\geq 0} c_n (A_{s,N} n + B_{s,N}) (2kk')^{2n}$$ with $$A_{s,N} = \frac{\sqrt{N}}{1+2s} (-k^2+k'^2) \qquad B_s(N) = \alpha_s(N) - \sqrt{N}k_N^2$$

If one uses instead the second series, and plug it into $(2)$: $$ 0 = -2\log(2kk') \frac{\cos \pi s}{\pi(1+2s)} + \sum_{n\geq 0} c_n (2kk')^{2n} (-A_{s,N} + (n A_{s,N} + B_{s,N}) d_n)$$

When $s=0$, this reduces to conjecture $(I)$ of OP: $d_n$ becomes $3\log 4 -6 \left(H_{2 n}-H_n\right)$; conjectures $(II), (III), (IV)$ comes from $s=1/6, 1/4, 1/3$ respectively. Q.E.D.

If one uses the expansion for $K'_s(k)^2 + K_s(k)^2$, one obtained similar series involving second order harmonic numbers.