Is there a way to rewrite a series like ∑n=1∞ as a power series? I know its

If |z|&lt;1, then 11−zn=1+zn+z2n+…⇒zn1−zn=zn+z2n+z3n+…

Since zn1−zn=∑m∈N:n|mzm you have: ∑n=1∞zn1−zn=∑m=1∞d(m)zm

If |z|&lt;1, then 11−zn=1+zn+z2n+…⇒zn1−zn=zn+z2n+z3n+…

Since zn1−zn=∑m∈N:n|mzm you have: ∑n=1∞zn1−zn=∑m=1∞d(m)zm

Resolved: "Is there a way to rewrite a series..."

Michael Maggard

Answered question

2021-12-22

Is there a way to rewrite a series like

\sum_{n = 1}^{\infty}

as a power series? I know its

Answer & Explanation

Lynne Trussell

Beginner2021-12-23Added 32 answers

| z | < 1

, then

\frac{1}{1 - z^{n}} = 1 + z^{n} + z^{2 n} + \dots \Rightarrow \frac{z^{n}}{1 - z^{n}} = z^{n} + z^{2 n} + z^{3 n} + \dots

limacarp4

Beginner2021-12-24Added 39 answers

We have

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

So,

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

Then,

\frac{z^{n}}{1 - z^{n}}

Now,

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

where

σ_{0} (n)

is the number of divisors of n, or, equivalently, the number of ways to write n as a product of two positive integers, order not mattering. Note that the number of times

(j + 1) k, j \geq 0, k j e q 1

equals n is given by this amount.

user_27qwe

Skilled2021-12-30Added 375 answers

Since $\frac{z^{n}}{1 - z^{n}} = \sum_{m \in N : n | m} z^{m}$ you have: $\sum_{n = 1}^{\infty} \frac{z^{n}}{1 - z^{n}} = \sum_{m = 1}^{\infty} d (m) z^{m}$

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