# Limit Comparison Test

Using the limit comparison test how do I find the test the convergence of the sum of $$\frac{1+2^{(n+1)}}{1+3^{(n+1)}}?$$

-
Do you need to use the limit comparison test? The ratio test may be more suited to seeing if the sum converges. – yunone Mar 25 '11 at 1:45
Or you could use the ordinary comparison test, that term being less than $\left(\frac{2}{3}\right)^n$. @Anora: Do you have any guesses as to what series to compare? – Jonas Meyer Mar 25 '11 at 2:05

Here, for large $n$ it should be clear that $2^{n+1}+1$ is "essentially" just $2^{n+1}$; and $3^{n+1}+1$ is "essentially" the same as $3^{n+1}$. So the fraction will be "essentially", for large $n$, about the same as $\frac{2^{n+1}}{3^{n+1}}$. So this suggests using limit comparison to compare $$\sum \frac{1+2^{n+1}}{1+3^{n+1}}$$ with $$\sum\frac{2^{n+1}}{3^{n+1}} = \sum\left(\frac{2}{3}\right)^{n+1}.$$ The latter is a geometric series, so it should be straightforward to determine whether it converges or not.
So let $a_n = \frac{1+2^{n+1}}{1+3^{n+1}}$, and $b_n = \left(\frac{3}{2}\right)^{n+1}$, and compute $$\lim_{n\to\infty}\frac{a_n}{b_n}.$$ If this limit exists and is positive (greater than $0$), then both series converge or both series diverge.