The sum of the series $$1 + \frac{1}{4 \cdot 2!} + \frac{1}{16 \cdot 4!} + \frac{1}{64 \cdot 6!} + \ldots$$ ad inf. is

The given infinite series is
$$S \;=\;1+\frac{1}{4\cdot2!}+\frac{1}{16\cdot4!}+\frac{1}{64\cdot6!}+\ldots$$

Write the general (⁠$$n^{\text{th}}$$⁠) term:

For $$n=0,1,2,\ldots$$ the denominator contains $$4^n$$ and $$(2n)!$$, so
$$T_n \;=\;\frac{1}{4^n\,(2n)!}$$

Hence
$$S \;=\;\sum_{n=0}^{\infty}\frac{1}{4^n\,(2n)!}$$

Notice that $$4^n=(2^2)^n=2^{2n}$$, therefore
$$\frac{1}{4^n}=\frac{1}{2^{2n}}=\left(\frac12\right)^{2n}$$

Substituting, we get
$$S \;=\;\sum_{n=0}^{\infty}\frac{\left(\dfrac12\right)^{2n}}{(2n)!}$$

Recall the Maclaurin expansion of the hyperbolic cosine:
$$\cosh x \;=\;\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$$

Comparing the two series, set $$x=\dfrac12$$. Then
$$S = \cosh\!\left(\frac12\right)$$

Evaluate $$\cosh\left(\dfrac12\right):$$
$$\cosh x = \frac{e^{x}+e^{-x}}{2}$$, so with $$x=\dfrac12$$,
$$S = \frac{e^{1/2}+e^{-1/2}}{2} = \frac{\sqrt{e}+\dfrac{1}{\sqrt{e}}}{2} = \frac{e+1}{2\sqrt{e}}$$

Therefore, the required sum is $$\dfrac{e+1}{2\sqrt{e}}$$.

Option D which is: $$\dfrac{e + 1}{2\sqrt{e}}$$