I am trying to show that the series $\sum _{n=1}^{\infty }\dfrac {nz^{n-1}\left( \left( 1+\dfrac {1} {n}\right) ^{n}-1\right) } {\left( z^{n}-1\right) \left( z^{n}-\left( 1+\dfrac {1} {n}\right) ^{n}\right) }$ converges absolutely for all values of $z$, except the values $z=\left( 1+\dfrac {a} {m}\right) e^{\dfrac {2k\pi i} {m}}$ ($a= 0, 1; k = 0,1,\ldots m-1;m=1,2,3,\ldots)$.
Since we are looking for absolute convergence D'Alembert's Ratio Test for absolute convergence and Gamma's convergence criterion come to mind. So if we can show $\lim _{n\rightarrow \infty }\left| \dfrac {U_{n+1}} {U_{n}}\right| =l < 1$ or $\lim _{n\rightarrow \infty }\left| \dfrac {U_{n+1}} {U_{n}}\right| = 1 +\dfrac {A_{1}} {n}+O\left( \dfrac {1} {n^{2}}\right) $, where $A_{1}$ is independent of n and $A_{1} < -1$, then we'll establish the series is absolutely convergent.
I was hoping to first establish that the series is absolutely convergent and wishfully thinking that i might stumble across an expression while doing this to prove the exception values.
Although this may seem like a good plan while solving the limit i am coming up against undefined expression as $n\rightarrow \infty $
I am unsure if i am going down the right line here any suggestion, alternative approaches , help would be much appreciated.
Edit: Maybe Cauchy's test if $\lim _{n\rightarrow \infty }\sup \left| U_{n}\right| ^{\dfrac {1} {n}} < 1$, then the series converges absolutely might be what we need here.