Question

Converge or Diverge? Summation n=1 to Inf of (1+4^n)/(1+3^n)

Answer 1

what is this limit

\[\Large\lim_{n\to\infty}\frac{1+4^n}{1+3^n}\]

Answer 2

Alternatively, you can use a series for comparison.
\[\frac{1+4^n}{1+3^n}\approx \left(\frac{4}{3}\right)^n\]
What do you know about the series,
\[\sum_{n=1}^\infty \left(\frac{4}{3}\right)^n\]

Answer 3

I broke the problem into separate fractions: 1/(1+3^n) and 4^n/(1+3^n) and the limit of the first is 0 and the second is infinity, that means it's divergent overall? I tried the comparison and it appears that (4/3)^n is a geometric series that diverges because |(4/3)|>1 and (1+4^n)/(1+3^n) <= (4/3)^n right? So that tells us nothing?

Answer 4

\[\Large\lim_{n\to\infty}\frac{1+4^n}{1+3^n}>\lim_{n\to\infty}\frac{4^n}{1+3^n}\]
\[\Large>\lim_{n\to\infty}\frac{4^n}{3^n+3^n}=\lim_{n\to\infty}\frac{1}{2}\cdot\frac{4^n}{3^n}\]
\[\large=\frac{1}{2}\lim_{n\to\infty}\left(\frac{4}{3}\right)^n=\infty\]