gvaldytist

2022-07-01

I need to find the limits of the following sequences
$\frac{\left(5n!+{n}^{9}-{2}^{n}\right)}{\left({2}^{n}+n!+{n}^{100}\right)}$

Sydnee Villegas

I will assume you want to evaluate
$\underset{n\to \mathrm{\infty }}{lim}\frac{5n!+{n}^{9}-{2}^{n}}{{2}^{n}+n!+{n}^{100}}.$
(This is just one sequence, of course.)
Divide everything by the "dominant term" (not the least dominant term). This is the factorial here:
$\begin{array}{}\text{(1)}& \frac{5n!+{n}^{9}-{2}^{n}}{{2}^{n}+n!+{n}^{100}}=\frac{5+\frac{{n}^{9}}{n!}-\frac{{2}^{n}}{n!}}{1+\frac{{2}^{n}}{n!}+\frac{{n}^{100}}{n!}}\end{array}$
From your comments concerning convergence and divergence, recall the definitions: the sequence $\left({a}_{n}\right)$ converges if $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}$ exists. In this case we say the sequence converges to the value of the limit. The sequence $\left({a}_{n}\right)$ diverges if $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}$ does not exist.
Informally a sequence converges if its terms get closer and closer to some number L as n gets larger and larger, and then we say the sequence converges to L. If the terms do not "settle down" to any number, the sequence diverges.
For your sequence, as n gets larger and larger, the non constant bits of the right hand side of (1) get closer and closer to 0. Then, the entire term on the right hand side of (1) gets closer and closer to 5. So the sequence converges to 5.

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