Infinite Factorial Summation-Part 2

Calculus Level 4

$\large \dfrac{3}{2!} + \dfrac{5}{4!} + \dfrac{7}{6!} + \ldots$

If the above series can be expressed as $S$ , find the value of $\big \lfloor 100S \rfloor$ .

1 solution

คลุง แจ็ค
Jun 19, 2015

First we have to find the formula for each term.

$\displaystyle \frac{3}{2!}+\frac{5}{4!}+\frac{7}{6!}+...=\Sigma_{n} \frac{2n+1}{(2n)!}$

$\displaystyle \Sigma_{n} (\frac{1}{(2n-1)!}+\frac{1}{(2n)!})$

$\displaystyle \Sigma_{n=1} \frac{1}{n!}=e-1$

$\displaystyle 100S=\boxed{171}$

ps. I can't type floor function.

isnt that e power -1?

Gokul Kumar - 5 years, 10 months ago

No, it's e-1

คลุง แจ็ค - 5 years, 10 months ago