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

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 $$...=\lim_{n\to\infty}(1+\frac{1}{n})^{n+2}=e$$

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

## 1 Answer

Correction:

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

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Wow ... so many careless mistakes ... in final exam revision ... – Jiew Meng Apr 20 '12 at 6:37