If ∑₍ᵣ₌₁₎ⁿ Tᵣ = ((2n − 1)(2n + 1)(2n + 3)(2n + 5)) / 64, then lim₍ₙ→∞₎ ∑₍ᵣ₌₁₎ⁿ (1 / Tᵣ) is equal to: (1) 1 (2) 0 (3) 2/3 (4) 1/3

1. Turning the cumulative sum into individual terms

The given expression is for the partial sum

$S_n\;=\;\sum_{r=1}^{n} T_r\;=\;\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$

To get just one term $T_n$ we use

$T_n\;=\;S_n\; -\; S_{n-1}$

This technique is standard in sequences & series: the difference of two successive partial sums equals the term that was added.

2. Doing the algebra cleanly

Rather than expanding the whole quartic, notice a pattern:

$S_n \;=\;\frac{[(2n-1)(2n+1)]\,[\,(2n+3)(2n+5)\,]}{64}$

We eventually find

$T_n=\frac{(2n-1)(2n+1)(2n+3)}{8}$

Key ideas students need:

• How to differentiate between term and sum of terms.
• Using $T_n=S_n-S_{n-1}$ to ‘pull out’ individual terms.
• Spotting factorisation patterns to avoid dirty expansions.

3. Reciprocals & partial fractions

Taking reciprocals gives

$\frac{1}{T_n}=\frac{8}{(2n-1)(2n+1)(2n+3)}$

A very common trick in IIT-JEE problems is turning such rational forms into telescoping sums. We look for constants $A,B\dots$ so that

$\frac{8}{(2n-1)(2n+1)(2n+3)}=\text{(something)}_{n}-\text{(something)}_{n+1}$

Partial-fraction decomposition reveals the compact identity

$\frac{8}{(2n-1)(2n+1)(2n+3)}=2\left[\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)}\right]$

Why choose this form? Because when you add successive $n$, every inner term cancels with the next, leaving only the first and last pieces—this is the definition of a telescoping series.

4. Summation and the limit

After cancellation the $n$-th partial sum of $\frac{1}{T_r}$ is

$\sum_{r=1}^{n}\frac{1}{T_r}= \frac{2}{3}-\frac{2}{(2n+1)(2n+3)}$

As $n\to\infty$ the leftover fraction $\frac{2}{(2n+1)(2n+3)}$ tends to zero, firmly planting the limit at $\frac{2}{3}$.

Concepts reinforced:

• Telescoping series — strategic partial-fractioning.
• Behaviour of rational terms as $n\to\infty$.
• Recognising that $p(n)\to0$ when degree of denominator $>0$ higher than numerator.

What is the question?

You are given the running total (also called the ‘sum till now’) of some mysterious numbers $T_1,\,T_2,\,\ldots$:

$\sum_{r=1}^{n} T_r=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$

The task is to find the final value of

$\lim_{n \to \infty} \;\sum_{r=1}^{n} \frac{1}{T_r}$

and then tick the correct option (1, 2, 3 or 4).

Kid-friendly road-map

1. Peel off one layer: To know $\frac{1}{T_r}$ we first need $T_r$. We will do that by seeing how much the running total jumps when we go from $n-1$ to $n$.
2. Find a neat formula: After a bit of algebra, $T_r$ turns out to be a simple product of three consecutive odd numbers divided by $8$.
3. Flip it: Taking reciprocals gives $\frac{1}{T_r}$.
4. Magic cancellation (telescoping): By writing $\frac{1}{T_r}$ as a ‘difference of two fractions’, almost everything cancels out when we start adding. Only the very first bit and the very last tiny bit survive!
5. Let $n$ go to infinity: The last tiny bit shrinks to zero, leaving a small fixed number behind.

Spoiler-free hint: the final leftover is two-thirds.

Step-by-step calculation

1. Define partial sums

$S_n=\sum_{r=1}^{n}T_r=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$

2. Extract $T_n$