Question

lf $a=\sum _{n=1}^{\infty }\frac{2n}{(2n-1)!},b=\sum _{n=1}^{\infty }\frac{2n}{(2n+1)!}$ then $ab$ is equal to

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

Answer 1

The correct option is A $1$

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

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

$\Rightarrow a=e$

$b=\sum _{n=0}^{\infty }\frac{(2n+1)-1}{(2n+1)!}$

$=\sum _{n=0}^{\infty }\frac{1}{\left(2n\right)!}-\frac{1}{(2n+1)!}$

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

So $ab=1$