Animated Solution for Mathematics - Sequence and Series: The sum of the series $\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\dots$ up to $\infty$ is equal to

Imagine you are standing at the edge of an infinite abyss, and before you lies a sequence of numbers, dancing in a rhythmic, alternating pattern. This is not just a list of fractions; it is a story of convergence, a puzzle where every term holds a secret.

We are looking at the series:

At first glance, it looks daunting, but let us break it down, step by step, with the precision of a master architect.

Every infinite series has a heartbeat, and that heartbeat is the general term, $T_n$. If we look at the denominators, we see $1\cdot 2$, $2\cdot 3$, $3\cdot 4$.

It is clear that the $n^{th}$ term has a denominator of $n(n+1)$. The signs alternate: positive, negative, positive, negative, which is the signature of $(-1{)}^{n-1}$.

Thus, our general term is:

This is the key that unlocks the door.

Now, here is where the JEE magic happens. Whenever you see consecutive integers multiplied in a denominator, your mind should immediately jump to partial fractions.

We can rewrite $\frac{1}{n(n+1)}$ as $\frac{1}{n}-\frac{1}{n+1}$. This simple algebraic manipulation transforms a complex product into a beautiful difference.

Now, our general term becomes:

Let us write out the first few terms to see the pattern emerge. For $n=1$, $T_1=(1-\frac{1}{2})$. For $n=2$, $T_2=-(\frac{1}{2}-\frac{1}{3})$. For $n=3$, $T_3=(\frac{1}{3}-\frac{1}{4})$.

When we sum these up, we get:

When we open the brackets, we must be careful with the signs. The expression becomes:

Notice the symmetry; the terms are grouping themselves.

We can group these terms as $S=1-2(\frac{1}{2})+2(\frac{1}{3})-2(\frac{1}{4})+\dots$. If we factor out a $2$, we get:

Now, this bracketed series looks familiar. It is the Maclaurin series for $\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots$.

If we set $x=1$, we get $\ln 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$. Our bracketed series is exactly $1-\ln 2$.

Substituting this back, we get:

Finally, using the properties of logarithms, $2\ln 2=\ln 4$ and $1=\ln e$.

The final result is:

And there it is—the elegant, final answer. You have conquered the infinite!