Question

Let $S$ denote the sum of the infinite series $1+\frac{8}{2!}+\frac{21}{3!}+\frac{40}{4!}+\frac{65}{5!} .......$. Then

• A. $S < 8$
• B. $S > 12$
• C. $8 < S < 12$
• D. $S = 8$

Answer 1

Answer: (C)

$8 < S < 12$

Answer 2

Let $S=1+\frac{8}{2 !}+\frac{21}{3 !}+\frac{40}{4 !}+\frac{65}{5 !}+\ldots$ Again, let $S_{1}=1+8+21+40+65+\ldots+T_{n}$ and $\frac{S_{1}=+1+8+21+40+\ldots+T_{n}}{0=1+7+13+19+25+\ldots-T_{n}}$ $T_{n}=1+7+13+19+25+\ldots+n$ terms $=\frac{n}{2}[2(1)+(n-1) 6]$ $=n[1+3(n-1)]=n(3 n-2)$ $\therefore S=\sum \frac{n(3 n-2)}{n !}$ $=\sum \frac{3 n-2}{(n-1) !}$ $=\sum \frac{3 n-3+1}{(n-1) !}$ $S=\sum \frac{3}{(n-2) !}+\sum \frac{1}{(n-1) !}$ $=3 e+e \,\,\,\,\left[\because e=1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots\right]$ $=4 e$ We know $2 < e < 3$ $\therefore 8 < 4 e < 12$ $\Rightarrow 8 < S < 12$