Prove divergence of series $1-\frac{1}{3}+\frac{2}{4}-\frac{1}{5}+\frac{2}{6}-\frac{1}{7}+\ldots$

If that series was convergent, then the series$$\left(1-\frac13\right)+\left(\frac12-\frac15\right)+\cdots+\left(\frac1n-\frac1{2n+1}\right)+\cdots$$would converge too. But$$\frac1n-\frac1{2n+1}=\frac{n+1}{2n^2+n}$$and you can use the comparison test (with respect to the harmonic series) to prove that the series $\displaystyle\sum_{n=1}^\infty\frac{n+1}{2n^2+n}$ diverges.

One more:

$1-1/3 +2/4-1/5+2/6-1/7+2/8.....=$

$(1/2+1/2-1/3)+ (1/4+1/4-1/5) + (1/6+1/6-1/7)+....\gt$

$(1/2 +1/3-1/3) +(1/4+1/5-1/5)+ (1/6+1/7-1/7)+..=$

$1/2+1/4+1/6+1/8+........=$

$(1/2)(1+1/2+1/3+1/4..........),$

harmonic series.

The Leibnitz test would not allow you to prove divergence, it is just a sufficient condition for convergence, not necessary. Your series can be written as

$$ \sum_{n=1}^{\infty}\left(\frac 1n -\frac{1}{2n+1} \right)=\sum_{n=1}^{\infty}\frac{n+1}{n(2n+1)}, $$

which is divergent by comparison with the harmonic series.