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{$\frac{3}{4^n} + \frac{2n}{3^n}$}
Here are the basic null sequences
(1) {$\frac{1}{n^p}$} for p>0;
(2) {$c^n$} for $|c| <1$
(3) {$n^pc^n$} for p>0 and $|c| <1$
(4) {$\frac{c^n}{n!}$} for any real c
(5) {$\frac{n^p}{n!}$} for p > 0
Using the second rule - I can see the denominators being null however I know I can't have 0 as a denominator. I can prove it's null using limits. However, the question specifically states to use the basic null sequences
Hint: Rewriting $$\frac{3}{4^n}+2\frac{n}{3^n}=3\cdot\left(\frac{1}{4}\right)^n+2\cdot n^1\left(\frac{1}{3}\right)^n$$ it is clear that the rule to be applied is $2$ for the first summand ($c=1/4<1$) and $3$ for the second.
