$$$\sum_{n=4}^{\infty} \frac{2^{- n} n!}{24 \left(n - 4\right)!}$$$

Déterminez $$$\sum_{n=4}^{\infty} \frac{2^{- n} n!}{24 \left(n - 4\right)!}$$$.

Simplify:

$$\sum_{n=4}^{\infty} \frac{2^{- n} n!}{24 \left(n - 4\right)!}=\sum_{n=4}^{\infty} \frac{2^{- n} n \left(n - 3\right) \left(n - 2\right) \left(n - 1\right)}{24}$$

Pull the constant out of the series:

$${\color{red}{\left(\sum_{n=4}^{\infty} \frac{2^{- n} n \left(n - 3\right) \left(n - 2\right) \left(n - 1\right)}{24}\right)}}={\color{red}{\left(\frac{\sum_{n=4}^{\infty} 2^{- n} n \left(n - 3\right) \left(n - 2\right) \left(n - 1\right)}{24}\right)}}$$

By the ratio test, the series is convergent.

Consider the sum $$$\sum_{n=4}^{\infty} z_{0}^{n}=\frac{z_{0}^{4}}{1 - z_{0}}$$$ (see steps).

By the ratio test, it is convergent when $$$\left|{z_{0}}\right| < 1$$$.

Differentiate both sides of the equality $$$4$$$ times with respect to $$$z_{0}$$$ (for steps, see derivative calculator):

$$\sum_{n=4}^{\infty} \frac{n z_{0}^{n} \left(n^{3} - 6 n^{2} + 11 n - 6\right)}{z_{0}^{4}}=- \frac{24}{z_{0}^{5} - 5 z_{0}^{4} + 10 z_{0}^{3} - 10 z_{0}^{2} + 5 z_{0} - 1}$$

Multiply both sides by $$$z_{0}^{4}$$$:

$$\sum_{n=4}^{\infty} n z_{0}^{n} \left(n^{3} - 6 n^{2} + 11 n - 6\right)=- \frac{24 z_{0}^{4}}{z_{0}^{5} - 5 z_{0}^{4} + 10 z_{0}^{3} - 10 z_{0}^{2} + 5 z_{0} - 1}$$

Setting $$$z_{0}=\frac{1}{2}$$$, we get that

$$\sum_{n=4}^{\infty} 2^{- n} n \left(n - 3\right) \left(n - 2\right) \left(n - 1\right)=48$$

Therefore,

$$\frac{{\color{red}{\left(\sum_{n=4}^{\infty} 2^{- n} n \left(n - 3\right) \left(n - 2\right) \left(n - 1\right)\right)}}}{24}=\frac{{\color{red}{\left(48\right)}}}{24}$$

Hence,

$$\sum_{n=4}^{\infty} \frac{2^{- n} n!}{24 \left(n - 4\right)!}=2$$

$$$\sum_{n=4}^{\infty} \frac{2^{- n} n!}{24 \left(n - 4\right)!} = 2$$$A