- #1
khari
- 13
- 0
Problem: Find the limit of the sequence
[tex]a_{n}=\frac{n^{2}2^{n}}{n!}[/tex]
My first thought was to say that
[tex]0\leq \frac{n^{2}2^{n}}{n!} \leq \frac{x^n}{n!} [/tex]
and by squeeze theorem, since [tex]\frac{x^n}{n!}=0[/tex] for all real x, my original limit must be 0 as well. Now all I need to do is prove that [tex]n^{2}2^{n} \leq x^n[/tex], which is where I'm stuck. Can anyone give me a hand? Thanks.
[tex]a_{n}=\frac{n^{2}2^{n}}{n!}[/tex]
My first thought was to say that
[tex]0\leq \frac{n^{2}2^{n}}{n!} \leq \frac{x^n}{n!} [/tex]
and by squeeze theorem, since [tex]\frac{x^n}{n!}=0[/tex] for all real x, my original limit must be 0 as well. Now all I need to do is prove that [tex]n^{2}2^{n} \leq x^n[/tex], which is where I'm stuck. Can anyone give me a hand? Thanks.
Last edited: