Animated Solution for Mathematics - Sequence and Series: The sum of the series $1+\frac{1}{4.2!}+\frac{1}{16.4!}+\frac{1}{64.6!}+\dots$ ad inf. is

Infinite series often feel like a labyrinth. You stare at a string of numbers, and it feels like a chaotic mess. However, in the world of JEE Advanced, every series is a puzzle waiting to be solved.

Today, we are going to unravel the mystery of the series:

At first glance, it looks intimidating, but let's break it down systematically.

Look closely at the denominators. We see $2!,4!,6!$—these are all even factorials. Right next to them, we have $4,16,64$.

If you have been practicing your calculus, you know that even factorials are a massive hint. They are the signature of the hyperbolic cosine function, $\cosh (x)$.

To derive this, we start with the fundamental exponential series:

When we add these two, the odd powers vanish, leaving us with:

This is the definition of $2\cosh (x)$.

Now, let's re-examine our series $S$. We observe that the denominators are powers of 4: $4^1,4^2,4^3,\dots$.

We can rewrite the terms as:

This simplifies to:

Do you see it now? This is exactly the $\cosh (x)$ series where $x=\frac{1}{2}$.

We have successfully collapsed the infinite series into a single, elegant function:

We know that $\cosh (x)=\frac{e^x+e^{-x}}{2}$. Plugging in $x=\frac{1}{2}$, we get:

Since $e^{1/2}=\sqrt{e}$ and $e^{-1/2}=\frac{1}{\sqrt{e}}$, our expression becomes:

To clean this up, we find a common denominator in the numerator:

And there it is! The infinite sum is simply $\frac{e+1}{2\sqrt{e}}$.

Isn't it beautiful how a complex-looking series simplifies into such a clean result? Keep this logic in your toolkit—whenever you see even factorials, think hyperbolic!