# Does the limit $\lim_{x\to1^{-}}(1-x)\sum_{n=1}^{\infty}2^{n} x^{2^n}$ exist?

Let the lacunary power series $$f(x):=(1-x)\sum_{n=1}^{\infty}2^{n} x^{2^n},0 We use Matlab to draw the function $$f(x)$$ graph, which shows that the function $$f(x)$$ has limits when $$x\to1^{-}$$. But we cannot rigorously prove it theoretically.

• You can also consider the continuous case Commented Aug 1 at 11:37
• @AdamRubinson. Compute $\int_{1}^{\infty}2^{n} \,x^{2^n}\,dn$ assuming $0<x<1$ Commented Aug 1 at 12:11
• math.stackexchange.com/a/3276562 $1/\ln(2)\approx1.44$ Commented Aug 1 at 12:27
• Hi @ClaudeLeibovici, as usual you are faster than I am. I can't see the direct relation of the integral in your comment and the sum in the OP. I hope I am not embarrassing myself here, but I have obtained the following: $$\frac12 \sum^\infty_{n=1}2^n x^{2^n} < \int^\infty_0 2^t x^{2^t}\,dt=\frac{x}{\log 2\log(x^{-1})} < 2\sum^n_{n=0}2^n x^{2^n}$$ but that does not give me (at least I don't see it) the existence of the limit $\lim_{x\rightarrow1-}(1-x)\sum^\infty_{n=1}2^n x^{2^n}$ Commented Aug 1 at 17:28
• @nelynx: That was what I initially did and found that the problem had more legs… Commented Aug 3 at 23:43