The series at hand is given by,
$$\sum_{k=1}^{\infty} \left(\frac{2k+1}{k (k+1)}\right)^s$$
I think it converges for $s>1$, but I have not been able to derive a general expression for this series.
I tried to evaluate it on mathematica and tried to find a pattern. This was an unsuccessful attempt. I also tried out a generating function approach, defining, $$F_s (x) = \sum_{k=1}^{\infty} \left(\frac{2k+1}{k (k+1)}\right)^s x^k$$ I started out with the series, $$\sum_{k=1}^{\infty} x^k$$ and found a series of appropriate steps involving taking derivatives and antiderivatives to obtain $F_1 (x)$, the idea being that the same steps must be applied recursively. In this way I wanted to find a pattern and attempt an inductive proof for $F_s (x)$ for any integer and I assumed that if I get it in terms of some 'nice' function (pardon the description, I am a physics student), the expression for any general $s$ would remain the same. But I hit a computational roadblock as I could not even find the expression for $s=2$.
I am looking for ideas on how to go about it.
Addendum: Would the closed form expression I find be the analytic continuation for the series for arbitrary $s$ in the vein of the Zeta function?
Partial solution
I don't think that the general closed form of the sum exists, though we can find the closed form in the specific case of $s=2k,\, k=1, 2, ...\,$, and also get interesting asymptotics at $s\to0$ and $s\to\infty$
Denoting $\displaystyle S_n=\sum_{k=1}^{\infty} \left(\frac{2k+1}{k (k+1)}\right)^n$, where $n=2, 4,...\,$,
we can write $$S_n=\frac12\sum_{k\neq0,1;\,k=-\infty}^\infty\left(\frac{2k+1}{k (k+1)}\right)^n$$ In a standart way we consider the integral in the complex plane along a big circle of the radius $R\to\infty$ $$I_R=\oint_{C_R}\pi\cot(\pi z)\left(\frac{2z+1}{z (z+1)}\right)^ndz\to 0\,\,\text{at}\,\,R\to\infty$$ On the other hand $$\lim_{R\to\infty}I_R=2\pi i\sum\underset{z=0;\pm1;\pm2...}{\operatorname{Res}}\,\pi\cot(\pi z)\left(\frac{2z+1}{z (z+1)}\right)^n$$ $$=2\pi i\left(S_n+\underset{z=0;\,-1}{\operatorname{Res}}\,\pi\cot(\pi z)\left(\frac{2z+1}{z (z+1)}\right)^n\right)=0$$ $$S_n=-\frac12\underset{z=0;\,-1}{\operatorname{Res}}\,\pi\cot(\pi z)\left(\frac{2z+1}{z (z+1)}\right)^n$$ As we have poles of order $n+1$ at $z=0;\,-1$ $$S_n=-\frac1{2(n!)}\left(\frac{d^n}{dz^n}\pi z\cot(\pi z)\left(\frac{2z+1}{z+1}\right)^n\,\bigg|_{z=0}+\frac{d^n}{dz^n}\pi (z+1)\cot(\pi z)\left(\frac{2z+1}{z}\right)^n\,\bigg|_{z=-1}\right)$$ $$=-\frac1{2(n!)}\frac{d^n}{dx^n}\pi x\cot(\pi x)\left(\left(1+\frac x{1+x}\right)^n+\left(1-\frac x{1-x}\right)^n\right)\,\bigg|_{x=0}$$ Using the decomposition of $\cot(\pi x)$ near $x=0$ ($B_n$ denotes the Bernoulli' number) $$\boxed{\,\,S_n=-\frac1{2(n!)}\frac{d^n}{dx^n}\,\bigg|_{x=0}\sum_{k=0}^\infty\frac{(-1)^k4^kB_{2k}}{(2k)!}(\pi x)^{2k}\left(\left(1+\frac x{1+x}\right)^n+\left(1-\frac x{1-x}\right)^n\right)\,\,}$$ This is the general answer for any even $n$.
Particularly, decomposing the expression near $x=0$
at $n=2$ $$S_2=-\frac1{2\cdot2!}\frac{d^2}{dx^2}\,\bigg|_{x=0}\left(1-\frac{(\pi x)^2}3\right)\big(2-2x^2\big)=1+\frac{\pi^2}3$$
at $n=4$ $$S_4=-\frac1{2\cdot4!}\frac{d^4}{dx^4}\,\bigg|_{x=0}\left(1-\frac{(\pi x)^2}3-\frac{(\pi x)^4}{45}\right)\big(2+4x^2+6x^4\big)=\frac{\pi^4}{45}+\frac{2\pi^2}3-3$$ etc.