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I was working on a problem when I encountered the infinite series $$\sum_{n=1}^{\infty} 2^{n}x^{\frac{1}{2^{n}}} = 2x^{\frac{1}{2}}+4x^{\frac{1}{4}}+8x^{\frac{1}{8}}+16x^{\frac{1}{16}}...$$ I've played around with it a bit but I am not sure how to approach this kind of infinite series. I don't know if there are common techniques to find a closed form for this type of infinite series(where the power of x itself is a geometric series), I only know basic geometric series and power series.

Thanks!

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    $\begingroup$ This does not converge for any non-zero $x$ (except when you have a $2$-adic topology). $\endgroup$
    – WhatsUp
    Commented Jul 3, 2020 at 21:17
  • $\begingroup$ Oh wow ya of course - I'm not sure how I missed that - thanks! $\endgroup$ Commented Jul 3, 2020 at 21:38

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For the lower bounds, you can compare the series to either of the two integrals (the former for $x>1$, the latte for $0<x<1$: $$ \int_{1}^{\infty}e^{\frac{1}{x}} dx\\ \int_{1}^{\infty}e^{-\frac{1}{x}}dx $$ both of which diverge

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