Calculus – Limits – JEE Main 29 Jan 2025 Shift 1

Solution:

Let $T_k = \frac{k^{3}+6k^{2}+11k+5}{(k+3)!}$.

We try to express the numerator in terms of factors of the denominator $(k+3)(k+2)(k+1) \dots$.

Observe that $(k+1)(k+2)(k+3) = (k^2+3k+2)(k+3) = k^3 + 3k^2 + 2k + 3k^2 + 9k + 6 = k^3 + 6k^2 + 11k + 6$.

The given numerator is $k^3 + 6k^2 + 11k + 5$.

So, Numerator $= (k^3 + 6k^2 + 11k + 6) – 1 = (k+1)(k+2)(k+3) – 1$.

Now substitute this back into $T_k$:

$T_k = \frac{(k+1)(k+2)(k+3) – 1}{(k+3)!}$

$T_k = \frac{(k+1)(k+2)(k+3)}{(k+3)!} – \frac{1}{(k+3)!}$

$T_k = \frac{1}{k!} – \frac{1}{(k+3)!}$.

This is a telescoping series. Let’s write out the sum $S_n = \sum_{k=1}^n T_k$:

$k=1: \frac{1}{1!} – \frac{1}{4!}$

$k=2: \frac{1}{2!} – \frac{1}{5!}$

$k=3: \frac{1}{3!} – \frac{1}{6!}$

$k=4: \frac{1}{4!} – \frac{1}{7!}$

…

Cancellation occurs between the negative term of $k=i$ and positive term of $k=i+3$.

The terms that remain are the first three positive terms and the last three negative terms.

$S_n = \left( \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} \right) – \left( \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \frac{1}{(n+3)!} \right)$.

Now take the limit as $n \to \infty$. The factorial terms in the second bracket tend to 0.

Limit $= 1 + \frac{1}{2} + \frac{1}{6}$.

Limit $= \frac{6+3+1}{6} = \frac{10}{6} = \frac{5}{3}$.

Ans. (4)