# Fungrim entry: 7f5468

$\sum_{n=1}^{\infty} \frac{\varphi\!\left(n\right) {q}^{n}}{1 - {q}^{n}} = \frac{q}{{\left(1 - q\right)}^{2}}$
Assumptions:$q \in \mathbb{C} \,\mathbin{\operatorname{and}}\, \left|q\right| < 1$
TeX:
\sum_{n=1}^{\infty} \frac{\varphi\!\left(n\right) {q}^{n}}{1 - {q}^{n}} = \frac{q}{{\left(1 - q\right)}^{2}}

q \in \mathbb{C} \,\mathbin{\operatorname{and}}\, \left|q\right| < 1
Definitions:
Fungrim symbol Notation Short description
Sum$\sum_{n} f\!\left(n\right)$ Sum
Totient$\varphi\!\left(n\right)$ Euler totient function
Pow${a}^{b}$ Power
Infinity$\infty$ Positive infinity
CC$\mathbb{C}$ Complex numbers
Abs$\left|z\right|$ Absolute value
Source code for this entry:
Entry(ID("7f5468"),
Formula(Equal(Sum(Div(Mul(Totient(n), Pow(q, n)), Sub(1, Pow(q, n))), For(n, 1, Infinity)), Div(q, Pow(Sub(1, q), 2)))),
Variables(q),
Assumptions(And(Element(q, CC), Less(Abs(q), 1))))

## Topics using this entry

Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub.

2019-09-22 15:43:45.410764 UTC