The sum of the series 1+142!+1164!+1646!+ ad infinitum is:

Question

The sum of the series $1+\dfrac{1}{4\cdot2!}+\dfrac{1}{16\cdot4!}+\dfrac{1}{64\cdot6!}+\cdots$ ad infinitum is:

Accepted Answer

$$\cosh x=\frac{e^x+e^{-x}}{2}=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\cdots$$

Putting $x=\dfrac12$: the given series $=\cosh\dfrac12=\dfrac{e^{1/2}+e^{-1/2}}{2}=\dfrac{\sqrt e+1/\sqrt e}{2}=\dfrac{e+1}{2\sqrt e}.$

Solutions · 1