A curious limit for $-\frac{\pi}{2}$ How to prove this ?
$$-\frac\pi2 = \lim_{x\to\infty}\sum_{n=1}^{\infty}(-1)^n \frac{x^{2n-1}}{(2n)! \ln 2n}$$
 A: Using 
$$\int_0^\infty \left(2 n\right)^{-t} \mathrm{d} t = \frac{1}{\ln(2n)}$$
the sum becomes
$$
  \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)! \cdot \ln(2n)} = \int_0^\infty \left(\sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)! \cdot (2n)^t} \right)\mathrm{d}t
$$
Now, further using 
$$
   \int_0^\infty u^{t-1} \mathrm{e}^{-2 n u} \mathrm{d} u = \Gamma(t) (2n)^{-t}
$$
we rewrite the sum as a double integral:
$$
   \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)! \cdot \ln(2n)} = \int_0^\infty \left( \int_0^\infty  \frac{u^{t-1}}{\Gamma(t)}\mathrm{d}t \right)  \frac{\cos\left(x \mathrm{e}^{-u}\right)-1}{x} \mathrm{d} u 
$$
In the large $x$ limit, the main contribution to the integral comes from large $u$. For large $u$, 
$$
    \int_0^\infty  \frac{u^{t-1}}{\Gamma(t)}\mathrm{d}t \approx \sum_{t=1}^\infty \frac{u^{t-1}}{\Gamma(t)} = \mathrm{e}^{u}
$$

Thus:
$$ \begin{eqnarray}
  \lim_{x \to \infty} \sum_{n=1}^\infty (-1)^n \frac{x^{2n-1}}{(2n)! \cdot \ln(2n)} &=& \lim_{x \to \infty} \int_0^\infty \mathrm{e}^{u} \frac{\cos\left(x \mathrm{e}^{-u}\right)-1}{x} \mathrm{d} u = \lim_{x \to \infty} \int_1^\infty \frac{\cos\left(x/w\right)-1}{x} \mathrm{d} w \\
  &=& \lim_{x \to \infty} \int_{1/x}^\infty \left(\cos\left(\frac{1}{v}\right)-1\right) \mathrm{d} v = \int_{0}^\infty \left(\cos\left(\frac{1}{v}\right)-1\right) \mathrm{d} v \\
  &=& -\frac{\pi}{2}
\end{eqnarray}
$$
