I've used the formula that tell us Murty from [1] in a special case to do calculations in the following
Example. I take the special case in page 211 of [1], taking the specialization $T=1/k$, where $k\geq 2$ is an integer, after I multiply the identity by $\mu(k)/k$, where $\mu(m)$ is the Möbius function, and taking the sum over all integers $k\geq 2$ one has $$\sum_{k=2}^\infty\frac{\mu(k)}{k}\sum_{n=2}^\infty\frac{(n+1)^2}{k^n}=\sum_{k=2}^\infty\mu(k)\frac{k(k+1)}{(k-1)^3}-4\sum_{k=2}^\infty\frac{\mu(k)}{k^2}-\sum_{k=2}^\infty\frac{\mu(k)}{k}.$$ Using the Dirichlet series $1/\zeta(s)$ for $\Re s>1$ where $\zeta(s)$ is the Riemann Zeta function, and the Prime Number Theorem one has by means of absolute convergence $$\sum_{n=2}^\infty(n+1)^2\left(\frac{1}{\zeta(n+1)}-1\right)=5-\frac{4}{\zeta(2)}+\sum_{k=2}^\infty\frac{\mu(k)k(k+1)}{(k-1)^3}.$$
I know that this example makes sense if one can justify the convergence of some of previous series.
When I returned to the main identity, first paragraph in page 211 of [1], and did previous tricks for the specilization $$T=\frac{1}{k^{\epsilon}},$$ for integers $k\geq 2$ and $\epsilon>\frac{1}{2}$ a real number (I believe that I can take here $z$ instead of $\epsilon$, as a complex number with $\Re z>\frac{1}{2}$, but I don't understand well the nature of the series in RHS) then I wrote as the series in LHS of my identity, that is the basis of my
Question. Justify the convergence of the series $$\sum_{n=2}^\infty \left( \left(\sum_{j=0}^n\alpha^{n-j}\beta^j\right)\left(\sum_{j=0}^n\gamma^{n-j}\delta^j\right)\cdot\left(\frac{1}{\zeta(\epsilon n+1)}-1\right) \right)$$ when $\epsilon>\frac{1}{2}$ is a real number (or tell us your extension for complex numbers, writing $z$ instead our previous $\epsilon$)? Many thanks.
[1] Ram Murty, Problems in Analytic Number Theory, Second Edition, Graduate Text in Mathematics 206, Springer (2008).
Lemma: Upper bound on $\zeta\left(1+x\right)$:
$$\zeta \left(1+x\right) = \frac{1}{1^{1+x}}+\frac{1}{2^{1+x}}+\frac{1}{3^{1+x}}+\frac{1}{4^{1+\frac{1}{x}}}+\frac{1}{5^{1+x}}+\cdots< \\ \frac{1}{1^{1+x}}+\frac{1}{2^{1+x}}+\frac{1}{2^{1+x}}+\frac{1}{4^{1+x}}+\frac{1}{4^{1+x}}+\cdots = \\ 1+\frac{2}{2^{1+x}}+\frac{4}{4^{1+x}}+\frac{8}{8^{1+x}}+\cdots = \\ 1+\frac{1}{2^{x}}+\frac{1}{4^{x}}+\frac{1}{8^{x}}+\cdots= \\ (2^{-x})^0+(2^{-x})^1+(2^{-x})^2+(2^{-x})^3+\cdots = \\ \frac{1}{1-2^{-x}}$$
This gives $\zeta \left(1+\varepsilon n\right) <\frac{1}{1-2^{-\varepsilon n}}$, so $\frac{1}{\zeta \left(1+\varepsilon n\right)} > 1-2^{-\varepsilon n}$, so $0 > \frac{1}{\zeta \left(1+\varepsilon n\right)}-1 > -2^{-\varepsilon n}$.
Because it is easier to deal with positive numbers, rewrite to $$0 < 1-\frac{1}{\zeta \left(1+\varepsilon n\right)} < 2^{-\varepsilon n}$$
Lemma: Now, we have $\left(\sum_{j=0}^n\alpha^{n-j}\beta^j\right) < \left(\sum_{j=0}^n\max\{\alpha,\beta\}^n\right) = (n+1)\max\{\alpha,\beta\}^n$. Similiar, $\left(\sum_{j=0}^n\gamma^{n-j}\delta^j\right) < \left(\sum_{j=0}^n\max\{\gamma,\delta\}^n\right) = (n+1)\max\{\gamma,\delta\}^n$. So if $$\max\{\alpha,\beta\}\max\{\gamma,\delta\}2^{-\varepsilon} < 1,$$
we can deduce that
$$0 < \sum_{n=2}^\infty \left( \left(\sum_{j=0}^n\alpha^{n-j}\beta^j\right)\left(\sum_{j=0}^n\gamma^{n-j}\delta^j\right)\cdot\left(1-\frac{1}{\zeta(\epsilon n+1)}\right) \right) < \sum_{n=2}^\infty \left( \max\{\alpha,\beta\}\max\{\gamma,\delta\}2^{-\varepsilon} \right)^n (n+1)^2 = \sum_{n=2}^\infty c^n(n+1)^2 $$
with $c<1$. Now let $c_n=c^n(n+1)^2$, then $\lim_{n \to \infty} \frac{c_{n+1}}{c_n} =\lim_{n \to \infty} \frac{c^{n+1}(n+2)^2}{c^n(n+1)^2} =c<1$, so by the ratio test it converges.
It can also be shown that if $\max\{\alpha,\beta\}\max\{\gamma,\delta\}2^{-\varepsilon} \geq 1$ the series diverge.
$$\zeta(1+x) > 1+ \frac{1}{2^{1+x}} = \frac{2^{1+x}+1}{2^{1+x}} $$
Therefore, $1-\frac{1}{\zeta(1+x)}>\frac{1}{2^{1+x}+1} > \frac{1}{2^{2+x}}$.
Therefore, $1-\frac{1}{\zeta(1+\varepsilon n)}>\frac{1}{2^{1+\varepsilon n}+1} > \frac{1}{2^{2+\varepsilon n}} = \frac14 2^{-\varepsilon n}$.
We can deduce that: $$\sum_{n=2}^\infty \left( \left(\sum_{j=0}^n\alpha^{n-j}\beta^j\right)\left(\sum_{j=0}^n\gamma^{n-j}\delta^j\right)\cdot\left(1-\frac{1}{\zeta(\epsilon n+1)}\right) \right) > \sum_{n=2}^\infty \left( \max\{\alpha,\beta\}\max\{\gamma,\delta\}2^{-\varepsilon} \right)^n \cdot \frac14 $$
which clearly diverges.