# Does a_n=(5^n)/(1+(6^n) converge? If so what is the limit?

Nov 10, 2015

Yes, it converges to zero.

#### Explanation:

First of all, I want to prove that $1 + {6}^{n}$ and ${6}^{n}$ are asymptotically equivalent. To do so, we need to show that

${\lim}_{n \setminus \to \setminus \infty} \frac{1 + {6}^{n}}{6} ^ n = 1$

And a way to show it is factoring ${6}^{n}$:

${\lim}_{n \setminus \to \setminus \infty} \frac{1 + {6}^{n}}{6} ^ n = {\lim}_{n \setminus \to \setminus \infty} \frac{\cancel{{6}^{n}} \left(1 + \frac{1}{6} ^ n\right)}{\cancel{{6}^{n}}} = {\lim}_{n \setminus \to \setminus \infty} 1 + \frac{1}{6} ^ n = 1$

Since the two expressions are equivalent when $n \to \infty$, we can claim that

${\lim}_{n \setminus \to \setminus \infty} {5}^{n} / \left(1 + {6}^{n}\right) = {\lim}_{n \setminus \to \setminus \infty} {5}^{n} / {6}^{n} = {\lim}_{n \setminus \to \setminus \infty} {\left(\frac{5}{6}\right)}^{n}$

And now use the fact that ${a}_{n} = {k}^{n}$ converges to zero if and only if $| k | < 1$, which is our case.