JEE Main 2025 — 29 January, Morning Shift. Previous Year Question.
Problem
The value of $\displaystyle\lim_{n\to\infty}\left(\sum_{k=1}^{n} \frac{k^3+6k^2+11k+5}{(k+3)!}\right)$ is:
Key insight. A sum with factorials in the denominator is a strong hint that it telescopes — but only once the numerator is rewritten in terms of the same shifted variable as the factorial. Substituting $j=k+3$ and decomposing $j^3$ using falling factorials ($j(j-1)(j-2)$, etc.) reveals the numerator collapses to just $j(j-1)(j-2)-1$, which splits neatly into $\frac{1}{k!}-\frac{1}{(k+3)!}$ — a textbook telescoping difference.
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Approach
Substitute $j=k+3$ so the factorial in the denominator becomes simply $j!$. Rewrite the numerator (originally in $k$) in terms of $j$, then decompose the resulting cubic using the identity $j^3=j(j-1)(j-2)+3j(j-1)+j$ to reveal it simplifies to $j(j-1)(j-2)-1$. Since $\frac{j(j-1)(j-2)}{j!}=\frac{1}{(j-3)!}$, this converts each term of the sum into $\frac{1}{k!}-\frac{1}{(k+3)!}$, which telescopes as $n\to\infty$.
Solution
Step 1 — Substitute j = k+3
Let $j=k+3$, so $k=j-3$ and the denominator becomes $j!$. Rewrite the numerator:
$$k^3+6k^2+11k+5 = (j-3)^3+6(j-3)^2+11(j-3)+5$$
Expanding each term and collecting:
$$(j-3)^3 = j^3-9j^2+27j-27, \quad 6(j-3)^2=6j^2-36j+54, \quad 11(j-3)=11j-33$$
Summing all terms plus the final $+5$:
$$j^3+(-9+6)j^2+(27-36+11)j+(-27+54-33+5) = j^3-3j^2+2j-1$$
Step 2 — Decompose j³ using falling factorials
Using the identity $j^3 = j(j-1)(j-2)+3j(j-1)+j$ (which expands to exactly $j^3$):
$$j^3-3j^2+2j-1 = \big[j(j-1)(j-2)+3j(j-1)+j\big] – 3\big[j(j-1)+j\big]+2j-1$$
(using $j^2=j(j-1)+j$)
$$= j(j-1)(j-2) + 3j(j-1)+j -3j(j-1)-3j+2j-1 = j(j-1)(j-2)-1$$
Step 3 — Rewrite the general term
$$\frac{j(j-1)(j-2)-1}{j!} = \frac{j(j-1)(j-2)}{j!} – \frac{1}{j!} = \frac{1}{(j-3)!}-\frac{1}{j!}$$
Substituting back $j=k+3$ (so $j-3=k$):
$$\frac{k^3+6k^2+11k+5}{(k+3)!} = \frac{1}{k!}-\frac{1}{(k+3)!}$$
Step 4 — Sum the telescoping series
$$\sum_{k=1}^{n}\left(\frac{1}{k!}-\frac{1}{(k+3)!}\right) = \sum_{k=1}^{n}\frac{1}{k!} – \sum_{k=1}^{n}\frac{1}{(k+3)!}$$
Shifting the index in the second sum ($m=k+3$, running from $4$ to $n+3$):
$$= \sum_{k=1}^{n}\frac{1}{k!} – \sum_{m=4}^{n+3}\frac{1}{m!}$$
Step 5 — Take the limit as n → ∞
As $n\to\infty$, both sums extend to infinity, and everything from $k=4$ onward cancels between the two sums, leaving only the first three terms of the first sum:
$$\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{1}{k!} – \sum_{m=4}^{n+3}\frac{1}{m!}\right) = \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}$$
Step 6 — Compute the final value
$$1+\frac{1}{2}+\frac{1}{6} = \frac{6+3+1}{6} = \frac{10}{6} = \frac{5}{3}$$
Answer
$$\frac{5}{3}$$
Common mistakes
- Trying to sum the series term by term for small $n$ and guessing the limit numerically. This can work for an estimate but risks missing the exact fraction; the telescoping decomposition gives the precise value directly.
- Getting the falling-factorial identity for $j^3$ wrong. The correct decomposition is $j^3=j(j-1)(j-2)+3j(j-1)+j$ — misremembering the coefficients (e.g. using $2j(j-1)$ instead of $3j(j-1)$) breaks the telescoping pattern entirely.
Practise next
- Evaluate $\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k^3+3k^2+2k}{(k+2)!}$, using the same falling-factorial decomposition and telescoping approach.
Show answer
$e$. The numerator factors as $k^3+3k^2+2k=k(k+1)(k+2)$, which is exactly the top three factors of $(k+2)!$.
$$\frac{k(k+1)(k+2)}{(k+2)!}=\frac{k(k+1)(k+2)}{(k+2)(k+1)k!}=\frac{k}{k!}=\frac{1}{(k-1)!}.$$
So the sum is $\displaystyle\sum_{k=1}^{\infty}\frac{1}{(k-1)!}=\sum_{j=0}^{\infty}\frac{1}{j!}=e$. Once the numerator is written as a falling factorial the whole thing collapses in one line.
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