Help simplifying this sum $f(x) =\sum_{n=1}^{\infty} \frac{2x}{n} e^{-x^2/n} 2^{-n}$, $ x \ge 0$

Question

I am stuck on this sum

$f(x) = \sum_{n=1}^{\infty} \frac{2x}{n} e^{-x^2/n} 2^{-n}$ $ x \ge 0$

Any tips on how to get started?

Thanks for any help

Answer 1

Perhaps this is a fruitful start. For $x\in\mathbb{R}$, notice that $$\begin{align} f(x) &= \sum_{n=1}^{\infty} \frac{2x}{n} e^{-x^2/n} 2^{-n} \\ &= -\sum_{n=1}^{\infty} 2^{-n} {d\over dx} e^{-x^2/n} \\ &= -F'(x) \end{align}$$ where $$ F(x)=\sum_{n=1}^{\infty} 2^{-n} e^{-x^2/n} $$ is absolutely convergent since each term is bounded above by $2^{-n}$ (and below by $2^{-n}e^{-x^2}$). In fact, it follows that $e^{-x^2}\le F(x)\le 1$ for all $x\in\mathbb{R}$, with equality iff $x=0$. The absolute convergence of $F$ on $\mathbb{R}$ allows us to differentiate term by term and bestows the same infinite radius of convergence on $f=-F'$.

The function $F$ is similar to a "mixed" distribution, a sum of Gaussian or normal distributions all centered at the origin, but with linearly increasing variance and geometrically decreasing weight. However the normalizing constants $(\pi n)^{-\frac12}$ are absent, (and the $x$-axis is contracted by the factor $\sqrt2$) so that the integral $$ \int_0^\infty F(x)dx =\frac{\sqrt\pi}2\sum_{n=1}^\infty{\sqrt{n}\over2^n}. $$ What was your goal? To prove that $f(x)$ converges, find its radius of convergence, or find an analytic formula for its sum?