TITLE: $\lim_{x\to 1^-}\sqrt{1-x}\ \left(1+x+x^4+x^9+x^{16}+x^{25}+\cdots\right)=\sqrt{\pi}/2$ is true?
QUESTION [19 upvotes]: When I was considering limits of various functions, I had the following conjecture.
$$\lim_{x \to 1^-}\sqrt{1-x}\ \sum_{k=0}^{\infty}x^{\left(k^2\right)}=\lim_{x\to 1^-}\sqrt{1-x}\ \left(1+x+x^4+x^9+x^{16}+x^{25}+\cdots\right)=\frac{\sqrt{\pi}}{2}$$
This seems true, but I can't prove that. If this is true, how can we prove that? If this is not true, what is the answer?

REPLY [6 votes]: We have the following theorem from G. Polya's book:

Theorem: Let for the monotonic function $f$ , $\int_{0}^\infty f(x)dx$ exists and we have $\lim_{x\to\infty}f(x)=0$ and $f(x)>0$ then
  we have 
$$\lim_{h\to0^+}h\sum_{v=0}^\infty f(vh)=\int_{0}^\infty f(x)dx.$$
By using this theorem we can show,
$$\lim_{t\to
 1^{-}}\sqrt[\alpha]{1-t}(1+t^{1^\alpha}+t^{2^\alpha}+t^{3^\alpha}+\cdots+t^{n^\alpha}+\cdots)=\frac{1}{\alpha}\Gamma(\frac{1}{\alpha}).$$

We need to take in previous theorem $t=e^{-h^{\alpha}}$ and $f(x)=e^{-x^\alpha}$ and since $\Gamma(\alpha)=\int_0^\infty e^{-x}x^{\alpha-1}dx$ and $\ln(1-t)\cong 1-t$.
See this book: G. Polya and Gabor Szegö, Problems and theorems in analysis , vol. I, Springer.