limit of $a_n= \frac{7^n + 6^n -n^{100}}{(7.1)^n-7^n+n^{101}}$ as n goes to infinity

Question

$a_n= \frac{7^n + 6^n -n^{100}}{(7.1)^n-7^n+n^{101}}$ consider the convergence of $a_n$

Answer 1

HINT

Recall that for any $a>1$ and $b$

$$\frac{n^b}{a^n}\to 0$$

then factor out the "stronger" term by numerator and denominator.

Answer 2

Try rewriting

$$a_n = \frac{(7.1)^n}{(7.1)^n}\frac{\left(\tfrac{7}{7.1}\right)^n+\left(\tfrac{6}{7.1}\right)^n-\tfrac{n^{100}}{7.1^n}}{1-\left(\tfrac{7}{7.1}\right)^n+\tfrac{n^{101}}{7.1^n}}$$

now what happens when $n\rightarrow \infty$

Answer 3

With equivalents:

$7^n + 6^n -n^{100}\sim_{n\to\infty}7^n$, $\quad(7.1)^n-7^n+n^{101}\sim_{n\to\infty}(7.1)^n$, so

$$\frac{7^n + 6^n -n^{100}}{(7.1)^n-7^n+n^{101}}\sim_{n\to\infty}\frac{7^n}{(7.1)^n}=\biggl(\frac 7{7.1}\biggr)^{\!n},$$ which is a geometric sequence with ratio $<1$.