Formal Limit Proofs for Limits Involving Factorials

Question

How does one use the definition of the limit to formally prove that

$$ \lim_{n \to \infty} \frac{n^4}{n^2 + n!} = 0? $$

See also Stirling's approximation. – Lucian Feb 25 '14 at 06:27 — Lucian, Feb 25 '14 at 06:27

score 3 · Answer 1 · answered Feb 25 '14 at 05:30

3

n! > n^5 for n > 10. Then 0 < n^4/(n^2 + n!) < 1/n and you can use squeeze theorem.

answered Feb 25 '14 at 05:30

DeepSea

77,651

score 2 · Answer 2 · answered Feb 25 '14 at 05:31

Well, you know that $n! \ge n(n-1)(n-2)(n-3)(n-4) \ge (n-4)^5$ (for $n >4$), so you have $0 \le{n^4 \over n^2+n!} \le {n^4 \over n!} \le {n^4 \over (n-4)^5} = {{1 \over n} \over (1-{4 \over n})^5 }$. Hence for $n>8$ we have $(1-{4 \over n}) \ge {1 \over 2}$, and so ${n^4 \over n^2+n!} \le 2^5 {1 \over n}$.

Now choose $n > \max(8, {2^5 \over \epsilon})$, then ${n^4 \over n^2+n!} < \epsilon$.

Formal Limit Proofs for Limits Involving Factorials

2 Answers2