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?
$$\lim_{x\to\infty}\sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2^k-1)x^k}{k k!}$$
 A: Hint: You're looking at $f(2x) - f(x) = \int_x^{2x} f'(t)\ dt$ where 
$f(z)  = \sum_{k=1}^\infty (-1)^{k+1} z^k/(k k!)$.  What is $f'(z)$?
A: $$\lim_{x\to\infty}\sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2^k-1)x^k}{k k!}=\lim_{x\to\infty}\sum_{k=1}^{\infty} \frac{(-x)^{k}}{k k!}-\sum_{k=1}^{\infty} \frac{(-2x)^{k}}{k k!}$$
Now $$f(x)=\sum_{k=1}^{\infty} \frac{x^{k}}{k k!}$$ fulfills 
$$f'(x)=\sum_{k=1}^{\infty} \frac{x^{k-1}}{k!}=\frac{e^x-1}{x}$$
So $$\lim_{x\to\infty}\sum_{k=1}^{\infty} \frac{(-x)^{k}}{k k!}-\sum_{k=1}^{\infty} \frac{(-2x)^{k}}{k k!}=\lim_{x\to\infty}f(-x)-f(-2x)=\lim_{x\to\infty}\int_{-2x}^{-x}\frac{e^t-1}{t} dt=\lim_{x\to\infty}\int_{-2x}^{-x}\frac{e^t}{t} dt-\lim_{x\to\infty}\int_{-2x}^{-x}\frac{1}{t} dt=0-\lim_{x\to\infty}(\log|-x|-\log|-2x|)=-\log(1/2)=\log(2)$$
A: 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
$$\lim_{x \to\infty} f(x) = \int_0^{\infty} dt \frac{e^{-t}-e^{-2 t}}{t} = \log{2}$$
