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Q. The sum of series $ \frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+... $ is

JamiaJamia 2007

Solution:

We know that $ e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+....\infty $ ?(i) $ {{e}^{-1}}=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-....\infty $ ?.(ii) On adding Eqs. (i) and (ii) $ e+{{e}^{-1}}=2+\frac{2}{2!}+\frac{2}{4!}+.....\infty $ $ \Rightarrow $ $ \frac{{{e}^{2}}+1}{e}-2=\frac{2}{2!}+\frac{2}{4!}+....\infty $ $ \Rightarrow $ $ \frac{{{e}^{2}}+1-2e}{e}=2\left[ \frac{1}{2!}+\frac{1}{4!}+....\infty \right] $ $ \Rightarrow $ $ \frac{{{(e-1)}^{2}}}{2e}=\frac{1}{2!}+\frac{1}{4!}+....\infty $