How to show that $\exists \lim_{x\rightarrow1^{-}}\sum_{n=0}^{+\infty} (-1)^n x^{n^2} = 1/2$?

Question

It is a well known fact that for $|x|<1$, we have $$\sum_{n=0}^{+\infty} (-1)^n x^n = \frac{1}{1+x}$$ How can we prove that $$\exists \lim_{x\rightarrow1^{-}}\sum_{n=0}^{+\infty} (-1)^n x^{n^2} = 1/2$$ and how would the results change if we replace $n^2$ by $n^k$ for some $k$?

It is not very difficult to observe that this is an alternating series and that it converges for $|x|<1$. The problem lies in the boundary, where the sum is divergent and so we can't apply Abel's theorem. Maybe there are some results in Tauberian theory but I don't know much about it.

Answer 1

You could use this with $\lambda_n:=n^2,\,s:=-\ln x$ so the limit is$$\lim_{R\to\infty}\frac{1}{(2N+1)^2}\sum_{n=0}^N(4n+1)=\frac12.$$In fact, this problem is the $k=2$ special case of the first comment under that answer.

Answer 2

I'll try to make David C. Ullrich's comment concrete. With $a(n^k)=(-1)^n, a(m)=0$, then $$\sum_{t\le N}\sum_{m\le t} a(m) = \sum_{(2n)^k \le N} (2n+1)^k-(2n)^k$$ $$= \sum_{(2n)^k \le N} k(2n)^{k-1}(1+O(n^{-1})=2^{k-1} (N^{1/k}/2)^k(1+o(1))=N 2^{-1}(1+o(1))$$

Thus, $$\lim_{x\to 1} \sum_{n\ge 0} (-1)^n x^{n^k}=\lim_{x\to 1}(1-x)\sum_{t\ge 0} (\sum_{m\le t} a(m)) x^t=\lim_{x\to 1}(1-x)^2\sum_{N\ge 0} (\sum_{t\le N}\sum_{m\le t} a(m)) x^N$$ $$=\lim_{x\to 1}(1-x)^2\sum_{N\ge 0} 2^{-1} N x^N (1+o(1))=\lim_{x\to 1}(1-x)^2 (\frac{x}{(1-x)^2}+o(\frac{x}{(1-x)^2}))=\frac12$$