Double

Quick answer: It should be clear that the numerator of the given expression is greater than the denominator, so we are looking for an answer $\gt 1$ . Only one option fill that requirement, namely $\dfrac{e^{2} + 1}{e^{2} - 1}$ .

Complete answer: Note first that $e^{x} = \displaystyle\sum_{n=0}^{\infty} \dfrac{x^{n}}{n!} = 1 + x + \dfrac{x^{2}}{2!} + \dfrac{x^{3}}{3!} + \dfrac{x^{4}}{4!} + ....$ , and so

$e^{-x} = \displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^{n} x^{n}}{n!} = 1 - x + \dfrac{x^{2}}{2!} - \dfrac{x^{3}}{3!} + \dfrac{x^{4}}{4!} - .....$ .

The numerator of the given expression is then equal to $\dfrac{e^{1} + e^{-1}}{2}$ , and the denominator to $\dfrac{e^{1} - e^{-1}}{2}$ .

The given expression can then be simplified to $\dfrac{e + \frac{1}{e}}{e - \frac{1}{e}} = \boxed{\dfrac{e^{2} + 1}{e^{2} - 1}} = \coth(1)$ .

Let us assume the value to be $y$ . $\frac{1+\frac{1}{2!}+\frac{1}{4!}+\cdots}{1+\frac{1}{3!}+\frac{1}{5!}+\cdots}=y$ Now applying componendo-dividendo, $\frac{\left(1+\frac{1}{2!}+\frac{1}{4!}+\cdots \right)+\left(1+\frac{1}{3!}+\frac{1}{5!}+\cdots \right)}{\left(1+\frac{1}{2!}+\frac{1}{4!}+\cdots \right)-\left(1+\frac{1}{3!}+\frac{1}{5!}+\cdots \right)}=\frac{y+1}{y-1}$ We can see that the numerator becomes $e$ and denominator becomes $e^{-1}$ , $\frac{e}{e^{-1}}=\frac{y+1}{y-1}$ $e^2=\frac{y+1}{y-1}$ $y=\boxed{\frac{e^2+1}{e^2-1}}$