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I am trying to show that the series $$\sum _{n=1}^{\infty }\dfrac {z^{n-1}} {\left( 1-z^{n}\right) \left( 1-z^{n+1}\right) }$$ is equal to $\frac {1} {\left( 1-z\right) ^{2}}$ when $\left|z\right| < 1$ and is equal to $\frac {1} {z\left( 1-z\right) ^{2}}$ when $\left|z\right| > 1$, so hence the series is not uniformly convergent around 1. This fact about not being uniformly convergent seems obvious here as the limit from the left and limit from the right and value at $z = 1$ are not the same, but i am unable to verify the sum of the series evaluates to those expressions under the conditions described.

I also observe that the partial sum of the series is $$\sum _{n=1}^{m }\dfrac {z^{m} -1} {\left( z-1\right) ^{2} \left( z^{m+1}-1\right) }$$ verified according to and when $\left|z\right| > 1$ $$\lim _{m\rightarrow \infty }\sum _{n=1}^{m }\dfrac {z^{m} -1} {\left( z-1\right) ^{2} \left( z^{m+1}-1\right) }=\dfrac {1} {z\left( 1-z\right) ^{2}}$$ checksout. I am unsure about the case when $\left|z\right| < 1$.

Any help would be much appreciated.

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@Arturo Thanks i shall keep that in mind. – Comic Book Guy Mar 21 '12 at 21:59
up vote 3 down vote accepted

I think you made a typo. Partial sums a given by the formula $$ \sum\limits_{n=1}^{m}\dfrac {z^{n-1}} {( 1-z^{n})( 1-z^{n+1})}= \frac{z^{m}-1}{( z-1)^{2}( z^{m+1}-1)} $$ If $|z|<1$, then $\lim\limits_{m\to\infty} z^m=0$, so $$ \sum\limits_{n=1}^{\infty}\frac{z^{n-1}}{( 1-z^{n})( 1-z^{n+1})}= \lim\limits_{m\to\infty}\sum\limits_{n=1}^{m}\frac{z^{n-1}}{( 1-z^{n})( 1-z^{n+1})}= \lim\limits_{m\to\infty}\frac{z^{m}-1}{( z-1)^{2}( z^{m+1}-1)}= $$ $$ \frac{0-1}{( z-1)^{2}( 0z-1)}=\frac{1}{(z-1)^{2}} $$ If $|z|>1$, then $\lim\limits_{m\to\infty} z^{-m}=0$, so $$ \sum\limits_{n=1}^{\infty}\frac{z^{n-1}}{( 1-z^{n})( 1-z^{n+1})}= \lim\limits_{m\to\infty}\sum\limits_{n=1}^{m}\frac{z^{n-1}}{( 1-z^{n})( 1-z^{n+1})}= \lim\limits_{m\to\infty}\frac{z^{m}-1}{( z-1)^{2}( z^{m+1}-1)}= $$ $$ \lim\limits_{m\to\infty}\frac{1-z^{-m}}{( z-1)^{2}(z-z^{-m})}= \frac{1-0}{(z-1)^{2}(z-0)}=\frac{1}{z(z-1)^{2}} $$

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thanks very much buddy. – Comic Book Guy Mar 21 '12 at 21:55

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