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Fungrim entry: 7f5468

n=1φ ⁣(n)qn1qn=q(1q)2\sum_{n=1}^{\infty} \frac{\varphi\!\left(n\right) {q}^{n}}{1 - {q}^{n}} = \frac{q}{{\left(1 - q\right)}^{2}}
Assumptions:qCandq<1q \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
Sumnf ⁣(n)\sum_{n} f\!\left(n\right) Sum
Totientφ ⁣(n)\varphi\!\left(n\right) Euler totient function
Powab{a}^{b} Power
Infinity\infty Positive infinity
CCC\mathbb{C} Complex numbers
Absz\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))))

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2019-09-22 15:43:45.410764 UTC