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I recently found this series from an Hardy work:

$$\sum_{n=0}^{+\infty}(-1)^nx^{2^n}=x-x^2+x^4-x^8+\dots$$

For what values of $x$ does it converge ? Can we use some summation technique to sum it where it should diverge ? For $x=1$ this looks like Grandi's series so I would say it's values should be $\frac 12$ or am I wrong ?

AlienRem
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  • ^ lol when x=1 it is Grandi's series, you can look it up here: https://en.wikipedia.org/wiki/Grandi%27s_series – Sorfosh Oct 15 '15 at 20:01
  • Yes I know about this but I'm wondering for other values like $x=2$ and so on... @Sorfosh – AlienRem Oct 15 '15 at 20:02
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    The series converges for $\lvert z\rvert < 1$. The unit circle is the natural boundary for the function, so there's no easy way to assign values to it for $\lvert x\rvert \geqslant 1$. Analytic continuation is right out, and the truly wild behaviour of the function near the unit circle doesn't bode too well for other techniques like Abel summation either. – Daniel Fischer Oct 15 '15 at 20:10
  • @DanielFischer well there is an easy way for x=1 and x=-1, right? – Sorfosh Oct 15 '15 at 20:14
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    @Sorfosh If you just "plug in" $\pm 1$ and ignore where things come from, then yes. If you keep track of where things come from, it's not so easy. – Daniel Fischer Oct 15 '15 at 20:48
  • Well, i was thinking of showing it is 1/2 form the taylors series of 1/x. Which seems quite simple :) @DanielFischer – Sorfosh Oct 15 '15 at 20:57

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Partial answer.

For $F_R(x)=Rx+R^2x^2+R^3x^4+R^4x^8+\cdots$, $R\neq1$, we have $F_R(x)-RF_R(x^2)=Rx$. Thus $F_R(1)$ belongs to the elementary Ramanujan class $R$ and is summable to $R/(1-R)$ (definition, also here). For $R=-1$, the elementary Ramanujan sum of the diluted Grandi series $1-1+0+1+0+0+0-1+\cdots$ is actually $1/2$.

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