Clues for $\lim_{x\to\infty}\sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2^k-1)x^k}{k k!}$ Some clues for this questions?
$$\lim_{x\to\infty}\sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2^k-1)x^k}{k k!}$$

Take the derivative and use the exponential series.  Thus if the sum is $f(x)$, then
$$x f'(x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{2^n-1}{n!} x^n = e^{-x}-e^{-2 x}$$
Then
$$f(x) = \int_0^x dt \frac{e^{-t}-e^{-2 t}}{t}$$
(because you know $f(0)=0$). Thus, using Fubini's theorem, one can show that
$$\lim_{x \to\infty} f(x) = \int_0^{\infty} dt \frac{e^{-t}-e^{-2 t}}{t} = \log{2}$$
