khari
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Problem: Find the limit of the sequence
a_{n}=\frac{n^{2}2^{n}}{n!}
My first thought was to say that
0\leq \frac{n^{2}2^{n}}{n!} \leq \frac{x^n}{n!}
and by squeeze theorem, since \frac{x^n}{n!}=0 for all real x, my original limit must be 0 as well. Now all I need to do is prove that n^{2}2^{n} \leq x^n, which is where I'm stuck. Can anyone give me a hand? Thanks.
a_{n}=\frac{n^{2}2^{n}}{n!}
My first thought was to say that
0\leq \frac{n^{2}2^{n}}{n!} \leq \frac{x^n}{n!}
and by squeeze theorem, since \frac{x^n}{n!}=0 for all real x, my original limit must be 0 as well. Now all I need to do is prove that n^{2}2^{n} \leq x^n, which is where I'm stuck. Can anyone give me a hand? Thanks.
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