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

To find the sum of the series, we use the Maclaurin expansion of the exponential functions.

Recall the standard infinite series expansion for $$ e^x $$:

$$ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots $$

Substitute $$ x = 1 $$ into the expansion formula:

$$ e^1 = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \ldots $$

Now substitute $$ x = -1 $$ into the expansion formula, which makes the signs alternate for odd powers:

$$ e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \ldots $$

Add the two equations together to eliminate all the odd factorial denominator terms:

$$ e^1 + e^{-1} = \left(1 + \frac{1}{1!} + \frac{1}{2!} + \ldots\right) + \left(1 - \frac{1}{1!} + \frac{1}{2!} - \ldots\right) $$

Combine the remaining identical terms on the right side of the expression:

$$ e + e^{-1} = 2(1) + 2\left(\frac{1}{2!}\right) + 2\left(\frac{1}{4!}\right) + 2\left(\frac{1}{6!}\right) + \ldots $$

Factor out the common multiplier 2 from the entire right side:

$$ e + e^{-1} = 2 \left(1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \ldots\right) $$

Divide both sides by 2 to isolate the grouped series terms:

$$ \frac{e + e^{-1}}{2} = 1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \ldots $$

Subtract 1 from both sides to isolate the exact target series asked in the problem:

$$ \left(\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \ldots\right) = \frac{e + e^{-1}}{2} - 1 $$

Find a common denominator on the right side to merge the terms:

$$ \frac{e + e^{-1}}{2} - \frac{2}{2} = \frac{e + e^{-1} - 2}{2} $$

Express the numerator fraction cleanly by converting the negative exponent term:

$$ \frac{e + \frac{1}{e} - 2}{2} = \frac{\frac{e^2 + 1 - 2e}{e}}{2} = \frac{e^2 - 2e + 1}{2e} $$

Recognize that the numerator forms a perfect square trinomial:

$$ e^2 - 2e + 1 = (e - 1)^2 $$

Substitute this back into the fraction to get the final simplified form:

$$ \frac{(e - 1)^2}{2e} $$

Final Answer:

The sum of the series is $$ \frac{(e - 1)^2}{2e} $$.