0
$\begingroup$

{$\frac{3}{4^n} + \frac{2n}{3^n}$}

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

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.