Convergence of \sum{\frac{2^n}{3^n - 1}} using the limit comparison test

[phosgene]

Homework Statement

Use the limit comparison test to determine whether the following series converges or diverges.

$\sum{\frac{2^n}{3^n - 1}}$

Where the sum is from n = 1 to n = ∞.

Homework Equations

The limit comparison test:

Suppose a_n>0 and bn>0 for all n. If the limit of a_n/b_n=c, where c>0, then the two series $\sum{a_{n}}$ and $\sum{b_{n}}$ either both converge or both diverge.

The Attempt at a Solution

So I get that the point is to find another series that you know diverges or converges that will also make the above limit easy to evaluate. All I can think to do is to compare it to something like $\sum{1/3^n}$ or $\sum{2^n}$, the first of which results in the limit not existing and the second of which results in the limit being zero. I think it's obvious that the series should converge, as the denominator is always equal to or greater than the numerator. But I'm out of ideas on what other sums I could use in the limit comparison test.

 

[LANS]

Compare it to
$$\sum \frac{2^n}{3^n}$$

$$\frac{2^n}{3^n}\cdot\frac{3^n-1}{2^n}$$
$$\frac{3^n}{3^n} - \frac{1}{3^n}$$
$$1- \frac{1}{3^n}$$
With limit n->infinity, this = 1

 

[DryRun]

Compare it to
$$\sum \frac{2^n}{3^n}$$

$$\frac{2^n}{3^n}\cdot\frac{3^n-1}{2^n}$$
$$\frac{3^n}{3^n} - \frac{1}{3^n}$$
$$1- \frac{1}{3^n}$$
With limit n->infinity, this = 1

Not quite, although the limit of the ratio is equal to 1. There are a few mistakes, in my opinion:

First, the ratio is wrong. It should be $u_n/v_n$ or in the OP's equation $a_n/b_n$. You've done the opposite. (A condition is missing in this case: c<∞)

Second, the limit is obtained using L'Hopital's rule.

Third, $\sum_{n=1}^{\infty} \frac{2^n}{3^n}$ is in fact a geometric series.

It should be easy now to form the final conclusion.

 

[phosgene]

Doh! That should've been my next try. I got too caught up trying to do the same thing I was doing for the preceding questions. Thanks for the help guys :)