The derivative of some expression turned out to be:
$$\frac{e^x}{x}(1 -\frac{1}{2x}) + e^x\sum_{n=2}^{\infty}\frac{x^{n -1}}{n!} + \frac12 \sum_{n =3}^{\infty}\frac{x^{n -2}}{n(n-2)!} + e^{-x}\sum_{n=2}^{\infty}\frac{(-x)^{n-1}}{n!} + \frac12 \sum_{n=3}^{\infty}\frac{(-x)^{n -2}}{n(n -2)!} + \frac{e^{-x}}{x} (1 + \frac{1}{2x})$$
But it should be $\frac{1}{x}$
Are these two expressions equal? I am currently exhausted and just looking at it became tiresome. Can someone help?
Thanks a lot.
No. $$\lim_{x \to \infty} \frac{e^x}{x} \left( 1 - \frac{2}{x} \right) = \infty \\[3ex] \lim_{x \to \infty} e^x \cdot \sum_{n=2}^{\infty} \frac{x^{n-1}}{n!} = \infty \\[3ex] \lim_{x \to \infty} \frac{1}{2} \sum_{n=3}^{\infty} \frac{x^{n-2}}{n(n-2)!} = \infty \\[3ex] \lim_{n \to \infty} e^{-x} \sum_{n=2}^{\infty} \frac{(-x)^{n-1}}{n!} = \lim_{x \to \infty} e^{-x} \cdot \frac{e^{-x} - 1 - (-x)}{-x} = 0 \\[3ex] \lim_{x \to \infty} \frac{e^{-x}}{x} \left( 1 + \frac{1}{2x} \right) = 0 $$
and
$$\lim_{x \to \infty} \frac{1}{2} \sum_{n=3}^{\infty} \frac{(-x)^{n-2}}{n(n-2)!} = \lim_{x \to \infty} \left[ -\frac{1}{2} \sum_{n=3}^{\infty} \frac{(-x)^{n-1}}{n!} \right]' = \lim_{x \to \infty} \left( \frac{1}{2x} \cdot \left[ e^{-x} - 1 - (-x) - \frac{(-x)^2}{2} \right] \right)' = \lim_{x \to \infty} \frac{2x \left( -e^{-x} + 1 - x \right) - 2 \left[ e^{-x} - 1 - (-x) - \frac{(-x)^2}{2} \right]}{4x^2} = -\frac{1}{4} $$
so the whole thing tends to $\infty$. But if it were $\frac{1}{x}$, it would tend to $0$.
