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In this questionthis question the OP asks whether the sum $$ f(q, \alpha) = \sum _{k=1}^{\infty } \frac{q^k \left(q^k-1\right)^\alpha}{(q;q)_k} $$ is ever zero. An experiment with Mathematica indicates, to any reasonable precision that $$f(2, 1) = \sum _{k=1}^{\infty } \frac{2^k }{(2;2)_{k-1}} = 0,$$ but I, for one, can't actually prove it. Is it true?

In this question the OP asks whether the sum $$ f(q, \alpha) = \sum _{k=1}^{\infty } \frac{q^k \left(q^k-1\right)^\alpha}{(q;q)_k} $$ is ever zero. An experiment with Mathematica indicates, to any reasonable precision that $$f(2, 1) = \sum _{k=1}^{\infty } \frac{2^k }{(2;2)_{k-1}} = 0,$$ but I, for one, can't actually prove it. Is it true?

In this question the OP asks whether the sum $$ f(q, \alpha) = \sum _{k=1}^{\infty } \frac{q^k \left(q^k-1\right)^\alpha}{(q;q)_k} $$ is ever zero. An experiment with Mathematica indicates, to any reasonable precision that $$f(2, 1) = \sum _{k=1}^{\infty } \frac{2^k }{(2;2)_{k-1}} = 0,$$ but I, for one, can't actually prove it. Is it true?

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GH from MO
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In this question the OP asks whether the sum $$ f(q, \alpha) = \sum _{k=1}^{\infty } (-1)^k\frac{q^k \left(q^k-1\right)^\alpha}{(q;q)_k} $$$$ f(q, \alpha) = \sum _{k=1}^{\infty } \frac{q^k \left(q^k-1\right)^\alpha}{(q;q)_k} $$ is ever zero. An experiment with Mathematica indicates, to any reasonable precision that $$f(2, 1) = \sum _{k=1}^{\infty } \frac{(-2)^k }{(2;2)_{k-1}} = 0,$$$$f(2, 1) = \sum _{k=1}^{\infty } \frac{2^k }{(2;2)_{k-1}} = 0,$$ but I, for one, can't actually prove it. Is it true?

In this question the OP asks whether the sum $$ f(q, \alpha) = \sum _{k=1}^{\infty } (-1)^k\frac{q^k \left(q^k-1\right)^\alpha}{(q;q)_k} $$ is ever zero. An experiment with Mathematica indicates, to any reasonable precision that $$f(2, 1) = \sum _{k=1}^{\infty } \frac{(-2)^k }{(2;2)_{k-1}} = 0,$$ but I, for one, can't actually prove it. Is it true?

In this question the OP asks whether the sum $$ f(q, \alpha) = \sum _{k=1}^{\infty } \frac{q^k \left(q^k-1\right)^\alpha}{(q;q)_k} $$ is ever zero. An experiment with Mathematica indicates, to any reasonable precision that $$f(2, 1) = \sum _{k=1}^{\infty } \frac{2^k }{(2;2)_{k-1}} = 0,$$ but I, for one, can't actually prove it. Is it true?

added 9 characters in body
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GH from MO
  • 105.4k
  • 8
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  • 398

In this question the OP asks whether the sum $$ f(q, \alpha) = \sum _{k=1}^{\infty } \frac{q^k \left(q^k-1\right)^\alpha}{(q;q)_k} $$$$ f(q, \alpha) = \sum _{k=1}^{\infty } (-1)^k\frac{q^k \left(q^k-1\right)^\alpha}{(q;q)_k} $$ is ever zero. An experiment with Mathematica indicates, to any reasonable precision that $$f(2, 1) = \sum _{k=1}^{\infty } \frac{2^k }{(2;2)_{k-1}} = 0,$$$$f(2, 1) = \sum _{k=1}^{\infty } \frac{(-2)^k }{(2;2)_{k-1}} = 0,$$ but I, for one, can't actually prove it. Is it true?

In this question the OP asks whether the sum $$ f(q, \alpha) = \sum _{k=1}^{\infty } \frac{q^k \left(q^k-1\right)^\alpha}{(q;q)_k} $$ is ever zero. An experiment with Mathematica indicates, to any reasonable precision that $$f(2, 1) = \sum _{k=1}^{\infty } \frac{2^k }{(2;2)_{k-1}} = 0,$$ but I, for one, can't actually prove it. Is it true?

In this question the OP asks whether the sum $$ f(q, \alpha) = \sum _{k=1}^{\infty } (-1)^k\frac{q^k \left(q^k-1\right)^\alpha}{(q;q)_k} $$ is ever zero. An experiment with Mathematica indicates, to any reasonable precision that $$f(2, 1) = \sum _{k=1}^{\infty } \frac{(-2)^k }{(2;2)_{k-1}} = 0,$$ but I, for one, can't actually prove it. Is it true?

fixed typo
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Igor Rivin
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Source Link
Igor Rivin
  • 96.4k
  • 11
  • 153
  • 366
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