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How fast does this grow: $$f(x) =\sum_{i=1}^{\infty} \dfrac{x^i}{i!^{1/2}}, \qquad x\in\mathbb{R}?$$

Is it faster than $e^{x^2}$? If Im not mistaken the Bell numbers grow fast enough such that $f(x)$ is at least slower then $e^{e^{x}-1}$.

Maybe solving $x^n = n!^{1/2}$ is relevant ?

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up vote 3 down vote accepted

When $x\to+\infty$, $f(x)$ behaves like $\mathrm e^{x^2/2}$ in the sense that $$ \lim_{x\to+\infty}\frac{\log f(x)}{x^2}=\frac12. $$

To prove this, fix $x\gt0$. For every $n\geqslant1$, $(2n+1)!\geqslant(2n)!\geqslant(2n-1)!$, hence $$ \frac{x^{2n-1}}{\sqrt{(2n-1)!}}+\frac{x^{2n}}{\sqrt{(2n)!}}\geqslant a_nx^{2n}(1+1/x), $$ and $$ \frac{x^{2n}}{\sqrt{(2n)!}}+\frac{x^{2n+1}}{\sqrt{(2n+1)!}}\leqslant a_nx^{2n}(1+x), $$ with $$ a_n=\frac1{\sqrt{(2n)!}}. $$ Summing these over $n\geqslant1$ yields $$ \sum\limits_{n\geqslant1}a_{n}x^{2n}(1+1/x)\leqslant f(x)\leqslant x+\sum\limits_{n\geqslant1}a_nx^{2n}(1+x). $$ For every $n\geqslant1$, $4^n(n!)^2\geqslant(2n)!\geqslant4^{n-1}((n-1)!)^2$, hence $$ \frac1{2^{n}(n!)}\leqslant a_n\leqslant\frac1{2^{n-1}(n-1)!}, $$ and $$ (1+1/x)(\mathrm e^{x^2/2}-1)\leqslant f(x)\leqslant x+(1+x)x^2\mathrm e^{x^2/2}. $$ This implies the result stated above.

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