The value of lim _(n → ∞)(Σₖ₌₁ⁿ (k³+6 k²+11 k+5)/(k+3)!) is :

Solution

Maths sir · 4 steps

Make the numerator look like (k+1)(k+2)(k+3), and each term splits into a difference of two factorials.

1. Spot the product. Expand (k+1)(k+2)(k+3) = k³+6k²+11k+6. The numerator is exactly this minus 1.

   (k³+6k²+11k+5)/(k+3)! = ((k+1)(k+2)(k+3)-1)/(k+3)! = 1/k!-1/(k+3)!

2. Telescope the sum. Adding from k=1 to n, every 1/m! with 4 ≤ m ≤ n appears once with a plus sign and once with a minus sign, so it cancels. Only three terms survive at the start and three at the end.
3. What is left. Sₙ = (1/1!+1/2!+1/3!)-(1/(n+1)!+1/(n+2)!+1/(n+3)!). As n → ∞, the second bracket goes to 0.
4. So. The limit is 1+1/2+1/6 = 5/3, which is option (a).

The idea to keep. When a sum has (k+r)! in the denominator, try writing the numerator as a product (k+1)(k+2)…(k+r) plus a constant. The product cancels the factorial and the sum telescopes.

Answer from the NTA final answer key. Solution by PrepWiser.