Question

If $a=\sum _{n=1}^{\infty }\frac{2n}{(2n-1)!},b=\sum _{n=1}^{\infty }\frac{2n}{(2n+1)!}$ , then $ab$ equals

• A. $1$
• B. $e^2$
• C. $\frac{e-1}{e+1}$
• D. $\frac{e+1}{e-1}$

Answer 1

The correct option is A $1$

$a=\sum _{n=1}^{\infty }\frac{2n}{(2n-1)!}$

$=\frac{2}{1!}+\frac{4}{3!}+\frac{6}{5!}+\frac{8}{7!}+------$

$=2+\frac{3+1}{3!}+\frac{5+1}{5!}+\frac{7+1}{7!}+-------$

$=1+(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+------)$

$=1+e-1=e\therefore a=e^1=e$

$a=\sum _{n=1}^{\infty }\frac{2n}{(2n+1)!}$

$=\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+\frac{8}{9!}+------$

$=\frac{3-1}{3!}+\frac{5-1}{5!}+\frac{7-1}{7!}+-------$

$=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}+--------$

$=1-1+\frac{1}{2}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+----------$

$b=e^{-1}\therefore b=e^{-1}$

$\therefore a\times b=e^1\times e^{-1}=e\times \frac{1}{e}=1$

$\therefore a\times b=1$