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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 $n^{2+n}$.2012-04-20

1 Answers 1

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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 ...2012-04-20