# Convergence of $\sum\limits_{m= 0}^{\infty} \frac{(-3)^m+2^m}{3^m+2^m}$

I'm trying to determine whether the series $$\sum\limits_{m= 0}^{\infty} \frac{(-3)^m+2^m}{3^m+2^m}$$ is absolutely or conditionally convergent or divergent. I've tried applying the comparison and ratio test to prove that it's not absolutely convergent without success. I've also been unable to apply the Leibniz test to prove that the series itself if divergent. My intuition tells me the series is divergent but I'm not sure how to prove it.

• For a series to converge, it is necessary for the general term to go to zero in the limit. What can you say about the general term of this series? In particular, what happens when $m$ is even? – Xander Henderson Mar 3 at 16:30
• @XanderHenderson Ah I see, thanks. – Hai Mar 3 at 16:33

## 1 Answer

Write the numerator as $$(-3)^m + (-2)^m -(-2)^m +2^m =(-1)^m((3)^m + (2)^m) +2^m(1-(-1)^m)$$.

The terms become $$(-1)^m+$$ a convergent sum, so the sum oscillates and is therefore divergent.