The harmonic numbers are defined by $$ H_n=\sum_{j=1}^n\frac{1}{j} $$ I have come across the following sum: $$ g(z)=\sum_{n=1}^\infty z^{H_n} $$ Clearly it converges only for $z<1/e$. Is it a known function?
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$\begingroup$ Here's a neat thing: $z^{H_n} = z^{1+1/2+1/3+...+1/n} = z\cdot z^{1/2}\cdot z^{1/3}\cdot\cdots z^{1/n}$ so, when it converges, the sum can be written as $g(z) = z(1+z^{1/2}(1+z^{1/3}(1+z^{1/4}(\cdots))))$. $\endgroup$– NealOct 4, 2016 at 12:58
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