Convergence of a sum using Leibniz test

Question

I tried to use ratio and root test to see the convergence of $$\sum_{n=1}^{+\infty}\frac{(-1)^{n}3n}{4n-1}$$ but both were inconclusive.

I also tried to use Leibniz test. I got that $|a_{n+1}|\leq|a_{n}|$, but I don't know how to calculate this limit: $$\lim_{n\to\infty}a_{n}$$ I know the series does not converge, but I can't show it.

Answer 1

Note that $$\lim_{n\to\infty}\frac {(-1)^n3n}{4n-1}$$

does not exist, since it can be rearranged to

$$\lim_{n\to\infty}(-1)^n\frac {3n}{4n-1}$$

We can then easily see that as $x$ approaches $\infty$, it oscillates between positive and negative values, while $\frac{3n}{4n-1}$ approaches $\frac{3}{4}$. Since it does not converge to 0, we can conclude that the series also diverges.

Answer 2

Hint: This series diverges since general term does not tend to zero.