How to evaluate $\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$? Is it possible to evaluate the sum:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$$
I expect it may be related to $\zeta^{\prime} (2)$:
$$\zeta^{\prime} (2) = - \sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$$
Is there an identity that works for my series, involving the natural logarithm, that is similar to the identity that:
$$\sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)} = \frac{\psi(a) - \psi(b)}{a-b}$$
Also potentially related, the Lüroth analogue of Khintchine’s constant can be defined as the following:
$$\sum_{n=1}^{\infty} \frac{\ln (n)}{n(n+1)}$$
as mentioned here.
After some work, the following can be shown:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{5\ln(2) + 4\ln(3)}{16} + \frac{1}{2} \sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right)$$
and furthermore:
$$\sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right) = \int_{0}^{2} \left( \frac{\left(1-\pi x \cot(\pi x) \right)}{2x^2} + \frac{1}{x^2 - 1} + \frac{1}{x^2 -4} \right) \, dx$$
EDIT
I have derived yet another form for my sum of interest, however, I found this one interesting as it seems like it could potentially be solvable?
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \int_{0}^{\infty} \left( \frac{\psi^{(0)} (s+3) + \gamma}{(s+2)(s-2)} - \frac{25}{16 (s-2)(s+1)} \right) \, ds$$
From this, it is possible to obtain the following:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4} i + \frac{25}{48} (\ln (2) - i \pi) - \frac{1}{8} + \frac{1}{16} i \pi + \frac{1}{4} \int_{0}^{i \pi} \psi^{(0)} \left( \frac{4}{1+ e^{u}} \right)  \, du$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4}i+\frac{25}{48} (\ln (2)-i \pi )+\frac{7 i \pi }{48}-\frac{1}{8}-\frac{\ln (2)}{3} -2 \int_0^{\infty } \frac{t \ln (\Gamma (1-i t))}{\left(t^2+4\right)^2} \, dt$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = -\frac{1}{8}-\frac{i \pi }{4}+\frac{i \gamma  \pi }{4}-\frac{\ln (2)}{16} - 2 \int_{0}^{\infty} \frac{t \ln (\Gamma (-i t)) }{(4+t^2)^2} \, dt$$
$$\implies \sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} =\frac{25}{48} \ln (2) -\frac{1}{8} + \int_{1}^{\infty} \frac{\ln (v-1) \text{li} (v^2)}{v^5} \, dv$$
Where $\text{li}$ is the logarithmic integral function.
$$\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2-4} = \frac{3 \ln (2)}{16} - \frac{\pi^2+1}{8} - \frac{\pi}{2} \int_{0}^{\infty} \sin(4\pi x) (\psi (x) - \ln (x)) \, dx$$
 A: Considering the more general case $$S_a=\sum_{k=a+1}^{\infty} \frac{\log(k)}{k^2 - a^2}$$
$$\frac{\log (k)}{k^2 - a^2}=\sum_{n=0}^\infty \frac{\log (k)}{k^{2n+2}} a^{2n}$$
$$\sum_{k=a+1}^{\infty} \frac{\log(k)}{k^{2n+2}}=-\text{HurwitzZeta}^{(1,0)}(2 n+2,a+1)$$
$$S_a=-\sum_{n=0}^\infty  a^{2n}\,\text{HurwitzZeta}^{(1,0)}(2 n+2,a+1)$$ converges very fast.
If the summation had started at $k=1$ instead of $k=a+1$, we would have obtained
$$-\sum_{n=0}^\infty  a^{2n}\,\zeta '(2 (n+1))$$
Considering the case of $a=2$ and the partial sums up to $p$, the value $0.920492$ is obtained for $p=17$.
$$\left(
\begin{array}{cc}
 p & -\sum_{n=0}^p \\
 1 & 0.866620 \\
 2 & 0.898968 \\
 3 & 0.911406 \\
 4 & 0.916559 \\
 5 & 0.918769 \\
 6 & 0.919732 \\
 7 & 0.920155 \\
 8 & 0.920343 \\
 9 & 0.920426 \\
 10 & 0.920462 \\
 11 & 0.920479 \\
 12 & 0.920486 \\
 13 & 0.920489 \\
 14 & 0.920491 \\
 15 & 0.920491 \\
 16 & 0.920491 \\
 17 & 0.920492
\end{array}
\right)$$ while the orginal summation requires tens of thousands.
This is explained by the fact that
$$\zeta '(2 (n+1)) \sim -\exp\Big[-\frac{1106 }{797}n-\frac{713}{430} \Big]$$
A: This is a partial answer and not very rigorous (I don't know how to make it rigorous) but i think this way might be able to produce some results.
$$ \sum_{k=3}^{\infty} \frac{\ln(k)}{k^2-4} - \sum_{k=3}^{\infty} \frac{\ln(k)}{k^2} = 4\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2(k^2-4)} $$
Now Consider the function
$$f(a) = \sum_{k=3}^{\infty} \frac{1}{k^a(k^2-4)}$$
such that :
$$\lim_{a \to 2}\, \frac{d}{da}\, f(a) = -\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2(k^2-4)} $$
Then using
$$ \int_0^1 x^{k-1} \, dx = \frac{1}{k} $$
One can show that if $a$ is an EVEN INTEGER greater than $0$
$$ \sum_{k=3}^{\infty} \frac{1}{k^a(k^2-4)} = \left.\left(\frac{1}{2^{a+2}x}+\frac{(4^{\frac{a}{2}+1}-3)x}{3*2^{a+2}}+\frac{(2a+1)x^2}{4*2^{a+2}} +\frac{\ln(1-x)}{2^{a+2}x^2}-\frac{x^2\ln(1-x)}{2^{a+2}}-\sum_{n=1}^{\frac{a}{2}}\frac{\operatorname{Li}_{(a-2n+2)}(x)}{2^{2n}} \right)\right\vert_{x=0}^{x=1} $$
Hopefully someone is able to produce some results from this , cheers.
Edit #1
$$ \sum_{k=3}^{\infty} \frac{1}{k^a(k^2-4)} = \frac{1}{3} +\frac{1}{2^{a+3}} + \frac{1}{2^{a+4}}+\frac{a}{2^{a+3}} - \sum_{n=1}^{\frac{a}{2}}\frac{\zeta{(a-2n+2)}}{2^{2n}} $$
Edit #2 (27/05/2021)
One last remark before i stop trying the problem. Once Again , not rigorous.
Consider the following identity for EVEN INTEGERS and $ x\ge 2$ which i will not prove but you can find in a similar manner here in result #20 :
https://mathworld.wolfram.com/InfiniteProduct.html
$$\prod_{n=3}^{\infty} \left(1-\frac{4}{n^x}\right) = \frac{1}{6\, \pi^{x/2} \left(1-\frac{1}{2^{x-2}}\right) i^{x/2-1}} \prod_{n=1}^{x/2} \sin\left(\pi \,2^{2/x} \,(-1)^{2n/x}\right) $$
Taking Natural Logs of both sides and differentiating with respect to x and some switching around we can obtain a neat little identity :
$$ -\frac{i \pi}{4} - \frac{4 \ln{(2)}}{2^x-4} - \frac{\ln(\pi)}{2} = 4 \sum_{n=3}^{\infty}\frac{\ln(n)}{n^x-4} - \frac{d}{dx} \sum_{n=1}^{x/2} \ln\left(\sin\left(\pi \,2^{2/x} \,(-1)^{2n/x}\right)\right) $$
Note : One can encounter similar sums of these forms the one with the $**\ln(\sin)**$ inside and these arise when dealing with Catalan constant ,$\zeta(3)$ and PolyGamma functions as far as i know, we simply don't understand their behavior yet. Providing further evidence that at least right now , the sum you asked for may not have a closed form.
A: This is not an answer, but its too long to be a comment. Also, this isn't a closed-form solution, but I think its an interesting approach. Also, I won't be rigorous here, I'll just be pretending everything converges nicely
We start with $$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$ with the goal of ending at
$$\sum_{n=3}^\infty \frac{\ln(k)}{k^2-4}$$
We will need k, k-2, and k+2 on the bottom, so I'll integrate to get those terms
$$\frac{1}{x}\left(\frac{1}{(1-x)}-1-x-x^2\right) = \sum_{n=3}^\infty x^{n-1}$$
$$\int\frac{1}{x}\left(\frac{1}{(1-x)}-1-x-x^2\right)dx = \sum_{n=3}^\infty \frac{x^{n}}{n}$$
$$\int x\int\frac{1}{x}\left(\frac{1}{(1-x)}-1-x-x^2\right) = \sum_{n=}^\infty \frac{x^{n+2}}{n(n+2)}$$
$$\int \frac{1}{x^5}\int x\int\frac{1}{x}\left(\frac{1}{(1-x)}-1-x-x^2\right) = \sum_{n=3}^\infty \frac{x^{n-2}}{n(n+2)(n-2)}$$
$$x^2\int \frac{1}{x^5}\int x\int\frac{1}{x}\left(\frac{1}{(1-x)}-1-x-x^2\right)$$
To clean things up, lets just allow that $$f(x)=x^2\int \frac{1}{x^5}\int x\int\frac{1}{x}\left(\frac{1}{(1-x)}-1-x-x^2\right)$$ and solve for it later.
Now notice that if we take the derivative and multiply by x we get a new n term. In particular
$$x\frac{d}{dx}f(x)=\sum_{n=3}^\infty \frac{nx^{n}}{n(n+2)(n-2)}$$
If we let that process be its own operation, with $D = x\frac{d}{dx}$, then applying that operation k times will give
$$D^k(f(x))=\sum_{n=3}^\infty \frac{n^{k}x^{n}}{n(n+2)(n-2)}$$
Then taking the derivative will respect to k gives.
$$\frac{d}{dk}D^k(f(x))=\sum_{n=3}^\infty \frac{\ln(n)n^{k}x^{n}}{n(n+2)(n-2)}$$
Then letting k=1, and x=1 will give
$$\frac{d}{dk}D^k(f(x))=\sum_{n=3}^\infty \frac{\ln(n)}{(n+2)(n-2)}$$

Simplification (I will fill this section out more if I decide to make this more than simply a sketch of how to get of the sum). For the initial sums, I have mainly used wolfram, because it is very tedious to compute otherwise. (Edit, left out a factor somewhere, so I'll need to edit the below calculations)
$$x\int_{0}^{x}\left(\frac{1}{1-t}-1-t-t^{2}\right)dt = \frac{x}{6}\left(-2x^{3}-3x^{2}-6x-6\ln\left(1-x\right)\right)$$
$$\frac{1}{x^5}\int_{0}^{x}\frac{t}{6}\left(-2t^{3}-3t^{2}-6t-6\ln\left(1-t\right)\right)dt = \frac{1}{120x^5}\left(x\left(-8x^{4}-15x^{3}-40x^{2}+30x+60\right)-60\left(x^{2}-1\right)\ln\left(1-x\right)\right)$$
$$x^2\int_0^x\left(\frac{1}{120t^5}\left(t\left(-8t^{4}-15t^{3}-40t^{2}+30t+60\right)-60\left(t^{2}-1\right)\ln\left(1-t\right)\right)\right)dt = -\frac{x^{2}\left(30\left(x^{2}-1\right)^{2}\ln\left(1-x\right)+x\left(16x^{4}-50x^{2}+15x+30\right)\right)}{240x^{4}} =f(x)$$
With that out of the way, I'll now look at how to define $D$ with fractional values.
Since we know we will only be taking one derivative, and then looking at the neighborhood around that one derivative, we can just take the first derivative and then add on the fractional part. So:
$$ x^k\frac{d^k}{dx^k} xf'(x)= \frac{x^k}{\Gamma(1-k)}\frac{d}{dx}\int_0^x\frac{f(t)}{(x-t)^k}dt  $$
Edit: I have defined D^k slightly wrong, I will edit this more when I have fixed the definition
Edit: I haven't yet checked this, but I believe this is the correct way to define D^k
We start with the fact that $D^k(x^n) = n^kx^n$ and extend this to any function which has a power series.
We have an easy way to get the natural powers of $n$ from using natural number applications of $D^k$ (i.e. $D^1$, $D^2$, etc. are all well-defined). We can use this craft $n^\alpha$ from these.
Notice that $$n^\alpha = \sum_{k=0}^{\infty} a_k n^k$$
Now, we just need to get the $n^k$ out of a function with a power series.
If $f(x) = \sum_{k=0}^{\infty} b_k x^k$, then applying $D^k$ gives
$$D^k(f(x)) = \sum_{n=0}^{\infty} b_n n^kx^n$$
Since we would like to just get the $n^k$ term, where $n=m$, then, we can do
$$\frac{D^k(f(x))}{b_mx^{m+1}} = \sum_{n=0}^{\infty} \frac{b_n}{b_m} n^kx^{(n-(m+1))}$$
Then take the residue, so
$$Res\left(\frac{D^k(f(x))}{b_mx^{m+1}}\right)=n^k$$
And so $$D^\alpha(f(x)) = \sum_{k=0}^{\infty} a_k Res\left(\frac{D^k(f(x))}{b_nx^{n+1}}\right)$$
A: Partial Answer
\begin{align}
\sum_{k=3}^\infty\frac{\ln k}{k^2-4}&=\sum_{k=3}^\infty\frac{\ln k}{k^2\left(1-\frac4{k^2}\right)}\\
&=\sum_{k=3}^\infty\frac{\ln k}{k^2}\sum_{l=0}^\infty\frac{2^{2l}}{k^{2l}}\\
&=\sum_{l=0}^\infty2^{2l}\sum_{k=3}^\infty\frac{\ln k}{k^{2l+2}}&&(\text{By Fubini's theorem})\\
&=\sum_{l=0}^\infty2^{2l}\left(-\zeta'(2l+2)-\frac{\ln2}{2^{2l+2}}\right)\\
&=-\sum_{l=0}^\infty\left(\frac{\ln2}{4}+\zeta'(2l+2)4^l\right)
\end{align}
And I couldn't proceed further. But as a side note, the convergence of this sum proves that
\begin{align}
&\lim_{l\to\infty}\frac{\zeta'(2l+2)2^{2l+2}}4=-\frac{\ln2}{4}\\
\implies &\lim_{x\to\infty}\zeta'(x)2^x=-\ln2
\end{align}
where the second step follows by the fact that Riemann zeta function is analytic.
A: There is a connection to $\zeta'$ at positive even values
Using partial fractions, we have
$$\frac{1}{k^{2n}(k^2-a)}=\frac{1}{a^n(k^2-a)}-\sum_{j=1}^{n}\frac{1}{a^{n-j+1}k^{2j}},$$
so that
$$S_r(n,a)=\sum_{k\ge r}\frac{\ln k}{k^{2n}(k^2-a)}=\frac{1}{a^n}\left(\sum_{k\ge r}\frac{\ln k}{k^2-a}-\sum_{j=1}^{n}a^{j-1}\sum_{k\ge r}\frac{\ln k}{k^{2j}}\right),$$
provided that $r>\sqrt a$. It is then easy to show that
$$\sum_{k\ge r}\frac{\ln k}{k^{2j}}=\zeta'(2j)+\sum_{k=2}^{r-1}\frac{\ln k}{k^{2j}},$$
with the $\zeta'(2j)$ terms having no simpler form that I'm aware of. As of now, I don't know how to deal with the sum $\sum_{k}\ln(k)/(k^2-a)$, but I will update once I do.
A: This is a partial answer which leaves one integral unevaluated. I'm not sure if it really helps or if it is just a reformulation replacing an unevaluated sum by an unevaluated integral.
We are looking for the sum
$$s = \sum_{k=3}^{\infty} \frac{\log(k)}{k^2-4}$$
Inserting
$$\log(k) = \int_0^1 \frac{x^{k-1}-1}{\log (x)} \, dx\tag{1}$$
and interchanging the order of summation and integration the integrand is given by the sum
$$i(x):=\frac{1}{\log (x)}\sum _{k=3}^{\infty } \frac{x^{k-1}-1}{k^2-4}\tag{2}$$
for which Mathematica gives
$$\begin{align}i(x)=\frac{4 x+2 x^2 -7 x^3 +x^4+4 \left(1-x^4\right) \log (1-x)}{16 x^3 \log (x)}\end{align}\tag{3}$$
now integrating gives the integral representation
$$s = s_i := \int_{0}^{1} i(x)\,dx= s(x)|_{x\to 1}-s(x)|_{x \to 0}\tag{4}$$
where antiderivative is given by
$$s(x) = \int i(x)\,dx = s_1(x) + s_2(x)\tag{5}$$
where
$$\begin{align}s_1(x) & = \int \frac{x^4-7 x^3+2 x^2+4 x}{16 x^3 \log (x)} \, dx\\
= & \frac{\text{Ei}(-\log (x))}{4}+\frac{1}{16} \text{Ei}(2 \log (x))-\frac{7 \text{li}(x)}{16}\\
& +\frac{1}{8} \log (\log (x))\end{align}\tag{6}$$
and
$$s_2(x) = \frac{1}{4}\int \frac{ \left(1-x^4\right) \log (1-x)}{ x^3 \log (x)} \, dx\tag{7}$$
Here $\text{Ei}(z)$ is the exponential integral. The integral $s_2$ remains unevaluated. And I am stuck here.
That the integral $s_{i}$ is convergent can be seen from the expansions at the endpoints of the integration interval.
We have
$$\begin{align}i(x \to 0)& \simeq -\frac{25}{48 \log (x)}+\frac{x^2}{5 \log (x)}\\
& +\frac{x^3}{12 \log (x)}+O\left(x^4\right)\end{align}\tag{8}$$
and
$$\begin{align}i(x \to 1) & \simeq -\frac{9}{16}+\frac{19 (x-1)}{32}\\
& +(x-2) \log (1-x) + O((1-x)^2)\end{align}\tag{9}$$
