Question

Prove that $\frac{4}{1!}+\frac{11}{2!}+\frac{22}{3!}+\frac{37}{4!}+\frac{56}{5!}+...\infty =6e-1$

Answer 1

$\frac{4}{1!}+\frac{11}{2!}+\frac{22}{3!}+\frac{37}{4!}+\frac{56}{5!}+.......$

$=\frac{4}{1!}+\frac{4+7}{2!}+\frac{4+7+11}{3!}+\frac{4+7+11+25}{4!}+\frac{4+7+11+25+19}{5!}+.......$

Now, the $n^{th}$ element in the series is

$t_n=\frac{(4+7+11+15+19+.....+n^{th}\text{element})}{n!}$

or, $t_n=\frac{4+\frac{(n-1)}{2}\{2\times 7+(n-2\left)4\right\}}{n!}$

or, $t_n=\frac{4+(n-1)(2n+3)}{n!}$

or, $t_n=\frac{(2n^2+n+1)}{n!}$

or, $t_n=\frac{\left(2n\right(n-1)+2n+n+1)}{n!}$

or, $t_n=2\frac{1}{(n-2)!}$ $+3\frac{1}{(n-1)!}+\frac{1}{n!}$.

Now the given series $=$$\sum _{n=1}^{\infty }{t}_n=$ $\sum _{n=2}^{\infty }$ $2\frac{1}{(n-2)!}$+$\sum _{n=1}^{\infty }$$3\frac{1}{(n-1)!}$ $+\sum _{n=1}^{\infty }$ $\frac{1}{n!}$ $=2e+3e+(e-1)=6e-1$.

[Since, $\sum _{n=1}^{\infty }\frac{1}{(n-1)!}$ $=e$ and $\sum _{n=1}^{\infty }\frac{1}{\left(n\right)!}$ $=e-1$.]