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

Question

Some clues for this questions?

$$\lim_{x\to\infty}\sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2^k-1)x^k}{k k!}$$

Answer 1

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 $f'(z)$?

Answer 2

$$\lim_{x\to\infty}\sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2^k-1)x^k}{k k!}=\lim_{x\to\infty}\sum_{k=1}^{\infty} \frac{(-x)^{k}}{k k!}-\sum_{k=1}^{\infty} \frac{(-2x)^{k}}{k k!}$$

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 $$\lim_{x\to\infty}\sum_{k=1}^{\infty} \frac{(-x)^{k}}{k k!}-\sum_{k=1}^{\infty} \frac{(-2x)^{k}}{k k!}=\lim_{x\to\infty}f(-x)-f(-2x)=\lim_{x\to\infty}\int_{-2x}^{-x}\frac{e^t-1}{t} dt=\lim_{x\to\infty}\int_{-2x}^{-x}\frac{e^t}{t} dt-\lim_{x\to\infty}\int_{-2x}^{-x}\frac{1}{t} dt=0-\lim_{x\to\infty}(\log|-x|-\log|-2x|)=-\log(1/2)=\log(2)$$

Answer 3

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

$$\lim_{x \to\infty} f(x) = \int_0^{\infty} dt \frac{e^{-t}-e^{-2 t}}{t} = \log{2}$$