Absolute convergence for all values except the values $z=\left( 1+\frac {a} {m}\right) e^{\frac {2k\pi i} {m}}$ 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 $$\lim _{n\rightarrow \infty }\left| \dfrac {U_{n+1}} {U_{n}}\right| =l &lt; 1$$ 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 $A_{1} &lt; -1$, 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 $$\lim _{n\rightarrow \infty }\sup \left| U_{n}\right| ^{\dfrac {1} {n}} &lt; 1$$, then the series converges absolutely might be what we need here.
First of all, the term 
$$
\left(1+\frac1n\right)^n-1
$$
approaches $e-1$ eventually, so it doesn't play any role in the convergence. The same can be said of 
$$
z^n-\left(1+\frac1n\right)^n
$$
when $|z|\leq1$, since $4\geq|z^n-(1+\frac1n)^n|\geq1$ eventually. So, for $|z|&lt;1$, the general term in your series of absolute values is comparable with $n|z|^{n-1}$, and the series converges. 

For $|z|&gt;1$, 
$$
\left|\dfrac {nz^{n-1} } {\left( z^{n}-1\right) \left( z^{n}-\left( 1+\dfrac {1} {n}\right) ^{n}\right) }\right|=\dfrac{n|z|^{n-1}}{|z|^{2n}|1-\frac1{z^n}|\,|1-\frac{(1+1/n)^n}{z^n}|}
$$
For $n$ big, the two differences in the denominator approach 1, so the convergence of the series is decided by 
$$
\frac{n|z|^{n-1}}{|z|^{2n}}=\frac{n}{|z|^{n+1}},
$$
Which shows that the series converges absolutely for every $z$ with $|z|&gt;1$. 

The case $|z|=1$: here $z=e^{2\pi i t}$, with $t\in [0,1]$. When $t$ is rational, the series is not defined (as there are infinitely many values of $n$ where $z^n=1$. For $t$ irrational the terms of the series are defined, but they do not go to zero as $n\to\infty$, so the series is not convergent.
In all honesty, the general term of the series as written has no desire to tend to $0$ for $|z|=1$, so what absolute convergence can we talk about?