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For complex $z$, show that the sum

$$\sum_{n = 1}^{\infty} \frac{z^{n - 1}}{(1 - z^n)(1 - z^{n + 1})}$$

converges to $\frac{1}{(1 - z)^2}$ for $|z| < 1$ and $\frac{1}{z(1 - z)^2}$ for $|z| > 1$.

Hint: Multiply and divide each term by $1 - z$, and do a partial fraction decomposition, getting a telescoping effect.

I tried following the hint, but got stuck on performing a partial fraction decomposition. After all, since all polynomials can be factored in $\mathbb{C}$, how do I know what the factors of an arbitrary term are? I tried writing

$$\frac{z^{n - 1}(1 - z)}{(1 - z^n)(1 - z^{n + 1})(1 - z)} = \frac{z^{n - 1}}{(1 - z)^3(1 + z + \dotsb + z^{n - 1})(1 + z + \dotsb + z^n)} - \frac{z^n}{(1 - z)^3(1 + z + \dotsb + z^{n - 1})(1 + z + \dotsb + z^n)},$$

but didn't see how this is helpful.

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HINT: Use $$ \frac{z^{n}-z^{n+1}}{(1-z^n)(1-z^{n+1})} = \frac{1}{1-z^n} - \frac{1}{1-z^{n+1}} $$

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  • $\begingroup$ That worked great! Thank you! I should point out that their 'hint' was misleading, as using this hint required multiplying and dividing by $z(1 - z)$ $\endgroup$ – tylerc0816 Dec 19 '12 at 22:09

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