Prove-that-for-all-complex-such-as-z-lt-1-n-1-z-n-z-n-1-2-n-1-nz-n-z-n-1-0-

Question Number 89937 by ~blr237~ last updated on 20/Apr/20

P r o v e t h a t f o r a l l c o m p l e x s u c h a s ∣ z ∣< 1 =

\underset{n = 1}{\sum^{\infty}} \frac{z^{n}}{{(z^{n} - 1)}^{2}} + \underset{n = 1}{\sum^{\infty}} \frac{{n z}^{n}}{z^{n} - 1} = 0

Commented by mathmax by abdo last updated on 20/Apr/20

w e h a v e f o r ∣ u ∣< 1 \sum_{p = 0}^{\infty} u^{p} = \frac{1}{1 - u} \Rightarrow \sum_{p = 1}^{\infty} {p u}^{p - 1} = \frac{1}{{(1 - u)}^{2}}

u = z^{n} \Rightarrow \frac{1}{{(1 - z^{n})}^{2}} = \sum_{p = 1}^{\infty} p {(z^{n})}^{p - 1} = \sum_{p = 1}^{\infty} p {(z^{p - 1})}^{n} \Rightarrow

\sum_{n = 1}^{\infty} \frac{z^{n}}{{(z^{n} - 1)}^{2}} = \sum_{n = 1}^{\infty} z^{n} \sum_{p = 1}^{\infty} p {(z^{p - 1})}^{n}

= \sum_{p = 1}^{\infty} p \sum_{n = 1}^{\infty} {(z . z^{p - 1})}^{n} = \sum_{p = 1}^{\infty} p \sum_{n = 1}^{\infty} z^{p n}

f r o m a n o t h e r s i d e

\sum_{n = 1}^{\infty} \frac{{n z}^{n}}{z^{n} - 1} = - \sum_{n = 1}^{\infty} {n z}^{n} \sum_{p = 0}^{\infty} z^{n p} = - \sum_{p = 1}^{\infty} p z^{p} \sum_{n = 0}^{\infty} z^{p n}

= - \sum_{p = 1}^{\infty} p \sum_{n = 0}^{\infty} z^{p (n + 1}) = - \sum_{p = 1}^{\infty} p \sum_{n = 1}^{\infty} z^{p n} \Rightarrow

\sum_{n = 1}^{\infty} \frac{z^{n}}{{(z^{n} - 1)}^{2}} + \sum_{n = 1}^{\infty} \frac{{n z}^{n}}{z^{n} - 1} = 0