Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Convergence of $$a_n=\frac{n^{2+n}}{n!}$$

I used the ratio test and have:

$$\lim_{n\to\infty} \frac{(n+1)^{3+n}}{(n+1)!}\cdot \frac{n!}{(n+1)^{2+n}} \\= \lim_{n\to\infty} \frac{(n+1)^{3+n}}{n+1}\cdot \frac{1}{(n+1)^{2+n}}\\= 1$$

Did I do something wrong? Correct answer appears to be $$...=\lim_{n\to\infty}(1+\frac{1}{n})^{n+2}=e$$

share|cite|improve this question
You have $(n+1)^{2+n}$ where you want $n^{2+n}$. – Gerry Myerson Apr 20 '12 at 4:59

1 Answer 1

up vote 4 down vote accepted


$$\lim_{n\to\infty} \frac{(n+1)^{3+n}}{(n+1)!}\cdot \frac{n!}{\color{Blue}n^{2+n}} $$

share|cite|improve this answer
Wow ... so many careless mistakes ... in final exam revision ... – Jiew Meng Apr 20 '12 at 6:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.