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Fungrim entry: 7f5468

n=1φ(n)qn1qn=q(1q)2\sum_{n=1}^{\infty} \frac{\varphi(n) {q}^{n}}{1 - {q}^{n}} = \frac{q}{{\left(1 - q\right)}^{2}}
Assumptions:qC  and  q<1q \in \mathbb{C} \;\mathbin{\operatorname{and}}\; \left|q\right| < 1
\sum_{n=1}^{\infty} \frac{\varphi(n) {q}^{n}}{1 - {q}^{n}} = \frac{q}{{\left(1 - q\right)}^{2}}

q \in \mathbb{C} \;\mathbin{\operatorname{and}}\; \left|q\right| < 1
Fungrim symbol Notation Short description
Sumnf(n)\sum_{n} f(n) Sum
Totientφ(n)\varphi(n) 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:
    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)))),
    Assumptions(And(Element(q, CC), Less(Abs(q), 1))))

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2021-03-15 19:12:00.328586 UTC