• # question_answer $\frac{\frac{1}{2\,!}+\frac{1}{4\,!}+\frac{1}{6\,!}+.....\infty }{1+\frac{1}{3\,!}+\frac{1}{5\,!}+\frac{1}{7\,!}+.....\infty }=$ A) $\frac{e+1}{e-1}$ B) $\frac{e-1}{e+1}$ C) $\frac{{{e}^{2}}+1}{{{e}^{2}}-1}$ D) $\frac{{{e}^{2}}-1}{{{e}^{2}}+1}$

Correct Answer: B

Solution :

$\frac{\frac{1}{2\ !}+\frac{1}{4\ !}+\frac{1}{6\ !}+......\infty }{1+\frac{1\ }{3\ !}+\frac{1}{5\ !}+\frac{1}{7\ !}+.......\infty }$ $=\,\frac{2\,\,\left[ \frac{1}{2\,!}+\frac{1}{4\,!}+\frac{1}{6\,!}+....\infty \right]}{2\,\,\,\left[ 1+\frac{1}{3\,!}+\frac{1}{5\,!}+\frac{1}{7\,!}+....\infty \right]}$ $=\frac{(e+{{e}^{-1}})-2}{(e-{{e}^{-1}})}=\frac{e+\frac{1}{e}-2}{e-\frac{1}{e}}=\frac{{{e}^{2}}+1-2e}{{{e}^{2}}-1}$ $=\frac{{{(e-1)}^{2}}}{(e-1)(e+1)}=\frac{e-1}{e+1}$.

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