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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.