Infinite Factorial Summation with Sum of the powers of 2

Calculus Level 4

$\large S= 1 + \dfrac{1+2}{2!} + \dfrac{1+2+2^2}{3!} + \dfrac{1+2+2^2+2^3}{4!} + \cdots$

Given that $S = Ae^2 + Be + C$ , where $A,B$ and $C$ are integers, and $\displaystyle e= \lim_{x\to 0} (1+x)^{1/x} \approx 2.718$ , find the value of $A\times B \times C$ .

Try more problems of this type here .

The answer is 0.

1 solution

Guilherme Niedu
May 9, 2016

From geometric progression formula:

$\large \displaystyle S = \sum_{n=0}^{\infty} \frac{2^{n+1} - 1}{(n+1)!}$

$\large \displaystyle S = \sum_{n=0}^{\infty} \frac{2^{n+1}}{(n+1)!} - \sum_{n=0}^{\infty} \frac{1}{(n+1)!}$

Making $m=n+1$ for both sums

$\large \displaystyle S = \sum_{m=1}^{\infty} \frac{2^{m}}{m!} - \sum_{m=1}^{\infty} \frac{1}{m!}$

$\large \displaystyle S = \left ( \sum_{m=0}^{\infty} \frac{2^{m}}{m!} - 1 \right) - \left ( \sum_{m=0}^{\infty} \frac{1}{m!} -1 \right )$

$\large \displaystyle S = (e^2 - 1) - (e-1)$

$\large \displaystyle S = e^2 - e$

Thus, $A = 1, B = -1, C = 0$ and their product is $\fbox{0}$

Infinite Factorial Summation with Sum of the powers of 2

Try more problems of this type here .

The answer is 0.

1 solution

0 pending reports