Animated Solution for Mathematics - Sequence and Series: If ${\sum }_{r=1}^n{T}_r=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$ then ${\lim }_{n\to \infty }{\sum }_{r=1}^n(\frac{1}{T_r})$ is equal to:

Welcome, future engineer! Today, we are going to dissect a problem that looks like a mountain of algebra but is actually a beautifully orchestrated dance of numbers.

We are given the sum of the first $n$ terms of a series:

We are asked to find the limit of the sum of the reciprocals of these terms. The first step is to stop looking at the sum and start looking for the individual terms.

The answer lies in the fundamental relationship $T_n=S_n-S_{n-1}$. If you know the total height of $n$ blocks and the total height of $n-1$ blocks, the difference between those two heights must be the height of the $n$-th block.

Now, let us calculate $S_{n-1}$ by replacing every $n$ with $n-1$:

Now, we subtract $S_{n-1}$ from $S_n$. Instead of expanding these massive products, we look for common factors:

The expression inside the bracket simplifies to $2n+5-2n+3=8$. Suddenly, the complexity collapses:

We are halfway there. The question asks for the sum of the reciprocals, ${\sum }_{r=1}^n\frac{1}{T_r}$. Flipping our expression for $T_n$, we get:

This is a classic setup for the Method of Differences. We note that the difference between the first and last factors in the denominator is $(2r+3)-(2r-1)=4$.

We rewrite the numerator $8$ as $2\cdot 4$ and replace the $4$ with the difference of the factors:

This is the heart of the telescoping series. When we sum this from $r=1$ to $n$, every intermediate term cancels out:

This simplifies to:

Finally, we take the limit as $n\to \infty$. The term $\frac{1}{(2n+1)(2n+3)}$ approaches zero, leaving us with:

And there you have it! A seemingly impossible problem solved by breaking it down, finding the pattern, and letting the algebra do the work.