What is the range of $y=\sqrt{1}x+\sqrt{2}x^2+\sqrt{3}x^3+...$? This is an infinite series, defined for $-1<x<1$. On desmos I entered the first 1000 terms and the range is from $-0.3786$$\cdots$ to infinity. I am trying to find the range (specifically the lower bound) of the infinite series.
I tried considering $y-xy$, but that did not help, because I cannot express $y-xy$ in closed form.
 A: Noting that
\begin{align*}
f(x) := \sum_{n=1}^{\infty} \sqrt{n} \, x^n
&= \frac{1}{\Gamma(1/2)} \sum_{n=1}^{\infty} \frac{\Gamma(1/2)}{\sqrt{n}} \, n x^n \\
&= \frac{1}{\Gamma(1/2)} \sum_{n=1}^{\infty} n x^n \int_{0}^{\infty} \frac{1}{\sqrt{t}} \, e^{-nt} \, \mathrm{d}t \\
&= \frac{1}{\Gamma(1/2)} \int_{0}^{\infty} \frac{x e^{-t}}{\sqrt{t} (1 - x e^{-t})^2} \, \mathrm{d}t.
\end{align*}
Now, since
$$ f'(x) = \frac{1}{\Gamma(1/2)} \int_{0}^{\infty} \frac{e^{-t}(1 + x e^{-t})}{\sqrt{t} (1 - x e^{-t})^3} \, \mathrm{d}t > 0 $$
for $-1 < x < 1$, it follows that $f(x)$ is strictly increasing for $-1 < x < 1$. Consequently,
$$ \inf_{x\in(-1,1)}f(x) = \lim_{x \to -1^-} f(x) = - \frac{1}{\Gamma(1/2)} \int_{0}^{\infty} \frac{e^{-t}}{\sqrt{t} (1 + e^{-t})^2} \, \mathrm{d}t. $$
Using the integral representation of the Dirichlet eta function, the last integral is equal to
$$ -\eta(-1/2) = (2^{3/2} - 1) \zeta(-1/2) \approx -0.3801048126. $$

Remark. We may represent $f(x)$ using the polylogarithm, i.e.,
$$f(x) = \operatorname{Li}_{-1/2}(x). $$
So, once we know that $f(x)$ is increasing on $(-1, 1)$, then the answer is simply $\operatorname{Li}_{-1/2}(-1)$, which reduces to the above answer. The purpose of this solution is to provide a self-contained argument showing that $f(x)$ is indeed increasing on $(-1, 1)$ and thus the infimum of $f$ over $(-1, 1)$ is achieved at the left endpoint of the interval.
