Evaluating the limit: $\lim\limits_{x \to \infty} \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{x^{2n-1}}{(2n)! \log (2n)}$

Solution 1:

EDITED. It suffices to prove the following claim.

Proposition. Assume that $f(z) = \sum_{n=2}^{\infty} a_n z^n$ has radius of convergence $R$ and define

$$ F(z) = \sum_{n=2}^{\infty} \frac{n a_n}{n-1} z^n = z \int_{0}^{z} \frac{f'(w)}{w} \, dw. $$

Then for all $|z| < R$, we have

$$ \sum_{n=2}^{\infty} \frac{a_n}{\log n} z^n = \int_{0}^{1} \int_{0}^{\infty} \frac{t^{s-1}}{\Gamma(s)} F(z e^{-t}) \, dtds. $$

Proof. The proof is a straightforward computation:

\begin{align*} \int_{0}^{1} \int_{0}^{\infty} \frac{t^{s-1}}{\Gamma(s)} F(z e^{-t}) \, dtds &= \sum_{n=2}^{\infty} \frac{n a_n}{n-1} z^n \int_{0}^{1} \int_{0}^{\infty} \frac{t^{s-1}}{\Gamma(s)} e^{-nt} \, dtds \\ &= \sum_{n=2}^{\infty} \frac{n a_n}{n-1} z^n \int_{0}^{1} \frac{ds}{n^s} \\ &= \sum_{n=2}^{\infty} \frac{a_n}{\log n} z^n. \end{align*}

In our case, we can set $f(z) = 1 - \cos z$ and consequently $F(z) = z \operatorname{Si}(z)$, where $\operatorname{Si}(z)$ is the sine integral. Then for real $x$, we have

\begin{align*} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n)!\log(2n)} x^{2n-1} &= \int_{0}^{1} \int_{0}^{\infty} \frac{t^{s-1}}{\Gamma(s)} e^{-t} \operatorname{Si}(x e^{-t}) \, dtds \\ &\hspace{3em} \xrightarrow[x\to\infty]{} \quad \frac{\pi}{2} \int_{0}^{1}\int_{0}^{\infty} \frac{t^{s-1}}{\Gamma(s)} e^{-t} \, dtds = \frac{\pi}{2} \end{align*}

where the limiting procedure can be easily justified by the dominated convergence theorem.

Similarly, setting $f(z) = e^{-z} - 1 + z$ gives $F(z) = z (\log z + \gamma + \Gamma(0,z))$ for $\operatorname{Re}(z) > 0$, where $\Gamma(s, z)$ is the upper incomplete gamma function. Plugging this back, we obtain

$$ \sum_{n=2}^{\infty} \frac{(-1)^{n}}{n!\log n} z^{n-1} = \log z + \gamma - \frac{1}{2} + \int_{0}^{1} \int_{0}^{\infty} \frac{t^{s-1}}{\Gamma(s)} e^{-t} \Gamma(0, z e^{-t}) \, dtds. $$

Note that the last integral vanishes if we let $z \to \infty$ along the cone $|\arg(z)| \leq \frac{\pi}{2} - \delta$. So this again confirms numerical observation made by Machinato.