$$\lim_{n\to \infty}\frac{(-1)^n\cdot 6^n-5^{1+n}}{5^n-(-1)^{n+1}\cdot 6^{n+1}}$$ How can I operate with $n$ in degree? Some sequences may be found using for example the property limit of $x^n$ at $n \to \infty$ if $|x|<1$ but can anything be applied to this?
Convergence Test-Calculating limits
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limits
convergence
1 Answers
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Hint: note that $\,5^{1+n}=5 \cdot 5^n\,$ in the numerator, $\,5^n = \frac{1}{5} \cdot 5^{n+1}\,$ and $-(-1)^{n+1}=(-1)^{n+2}$ in the denominator, then rewrite the fraction as:
$$\frac{(-1)^n\cdot 6^n-5^{1+n}}{5^n-(-1)^{n+1}\cdot 6^{n+1}} = \frac{6^n}{6^{n+1}} \, \frac{(-1)^n - 5\cdot\left(\cfrac{5}{6}\right)^n}{(-1)^{n+2} + \cfrac{1}{5}\cdot\left(\cfrac{5}{6}\right)^{n+1}}$$
Finally, note that $\lim_{n\to \infty} \left(\cfrac{5}{6}\right)^n = 0\,$ and $\,(-1)^{n+2}=(-1)^n\,$.