$1 + \frac{{\log_e x}}{{1!}} + \frac{{(\log_e x)^2}}{{2!}} + \frac{{(\log_e x)^3}}{{3!}} + \dots \infty = $

$\frac{1}{1!} + \frac{4}{2!} + \frac{7}{3!} + \frac{10}{4!} + \dots \infty = $

$\sum_{k=1}^{\infty} \frac{1}{k !} \left(\sum_{n=1}^k 2^{n-1}\right)$ is equal to

$\frac{2}{1!}(\log_e 2) + \frac{2^2}{2!}(\log_e 2)^2 + \frac{2^3}{3!}(\log_e 2)^3 + \dots \infty = $

Let $S_{n} = 1 \cdot (n-1) + 2 \cdot (n-2) + 3 \cdot (n-3) + \dots + (n-1) \cdot 1$,for $n \geq 4$. The sum $\sum_{n=4}^{\infty} \left( \frac{2 S_{n}}{n!} - \frac{1}{(n-2)!} \right)$ is equal to:

The sum to infinity of the series $1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots$ is