Finding convergence or divergence of series $\sum^{\infty}_{k=0}(-1)^{k}\frac{k}{3k-1}$

Question

Finding convergence or divergence of series $$\sum^{\infty}_{k=0}(-1)^{k}\frac{k}{3k-1}$$

What i try:I am trying to solve it using Leibniz test

Let $\displaystyle a_{k}=\frac{k}{3k-1}$. Then $$a_{k+1}<a_{k}$$

And $$\lim_{k\rightarrow \infty}\frac{k}{3k-1}=\frac{1}{3}(\neq 0)$$

So leibniz test failed here.

Can anyone please explain me How do i find convergency of that series. Thanks

Answer 1

It is divergent, because of found by you: general member of series should tend to zero.