Prove divergence of series $1-\frac{1}{3}+\frac{2}{4}-\frac{1}{5}+\frac{2}{6}-\frac{1}{7}+\ldots$ 
Prove the divergence of the series: 
  $$  1-{1\over3}+{2\over4}-{1\over5}+{2\over6}-{1\over7}+\ldots$$

Attempt. Of course the test of Leibniz for alternating series does not apply, since the terms $1,1/3,2/4,...$ are not decreasing (besides, it would imply the convergence of the series, which is not our case). I thought of working on the partial sums $(s_n)$, especially
$$s_{2n}=1-{1\over3}+{2\over4}-{1\over5}+{2\over6}-{1\over7}+\ldots
{2\over2n}-{1\over 2n+1}$$ in order to prove divergence, but i didn't manage to do so.
Thanks in advance for the help.
 A: 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.
A: 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.
A: 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.
A: Your series
$$
\frac 22- \frac 13 + \frac 24 - \frac 15 + \frac 26 + \ldots
$$
is the sum of the convergent alternating series
$$
\frac 12- \frac 13 + \frac 14 - \frac 15 + \frac 16 + \ldots
$$
and the divergent series
$$
 \frac 12 \left( 1 + 0  + \frac 12 + 0 + \frac 13 + \ldots \right)
$$
and therefore divergent.
A: Your series can seriously be rewritten as
$$s=1+\sum_{n=1}^{\infty}(-1)^n\cdot\frac{3+(-1)^n}{2n+4}.$$
This is the first important step to avoid ambiguity.
Considering $N$-th partial sum of the infinite series, we deduce
$$\sum_{n=1}^{N}(-1)^n\cdot\frac{3+(-1)^n}{2n+4}=\sum_{n=1}^{N}\frac{3\cdot (-1)^n}{2n+4}+\sum_{n=1}^{N}\frac{1}{2n+4}.$$
Whereas the first partial sum converges to some finite value (simply applying Leibniz' criterion), the second one tends to $+\infty$ by comparision with harmonic sum.
Hence, the given infinite series is divergent.
A: Your series is 
$$\sum_{n \geq 1} \frac{1+3(-1)^n}{2(n+1)} $$
Hence the general term is the sum of $$\frac{3(-1)^n}{2(n+1)}$$ which is the general term of a convergent series (by Leibniz rule), and $$\frac{1}{2(n+1)}$$ which is the general term of a divergent series.
Therefore the series is divergent.
