7

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.

Lalit Tolani
  • 3,426
Dan
  • 22,158
  • 2
    Given that it's a convergent series in that interval, it's a continuous function, hence it's range is also an interval (as the domain is an interval). Thus, we are reduced to finding the endpoints. However, the right endpoint should be $+\infty$ (might require some proof, but roughly the reason is that the series can't be defined at $1$, and if the right endpoint were not infinity then it would have been definable) , which is why locating the left endpoint finishes the problem. (This clarifies the "specifically" : if you get the lower bound, you are done). – Sarvesh Ravichandran Iyer Aug 07 '21 at 06:30
  • As for the function itself, the partial sum requires non-elementary functions, see here. Of course that doesn't rule out having discernible values at fixed inputs. – Sarvesh Ravichandran Iyer Aug 07 '21 at 06:34
  • 2
    According to Mathematica, the answer is $(2\sqrt{2}-1)\zeta(-1/2)$. See: polylogarithm function. – YiFan Tey Aug 07 '21 at 06:35

1 Answers1

10

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.

Sangchul Lee
  • 167,468
  • Nice solution and (+1) for sure. We could do the same starting from $$\sum_{k=1}^n \sqrt k ,,x^k=\text{Li}_{-\frac{1}{2}}(x)-x^{n+1} \Phi \left(x,-\frac{1}{2},n+1\right)$$ – Claude Leibovici Aug 07 '21 at 08:16