Proving whether a series converges (Wolfram Alpha says it diverges)

Question

I was given the sequence$$c_j=(-1)^j\frac{2}{j+2}\left(1+\frac{1}{2}+\dots+\frac{1}{j+1}\right)$$ and I rewrote that as $$c_j=(-1)^j\frac{2}{j+2}\sum_{n=1}^{j+1}\frac{1}{n}=\sum_{n=1}^{j+1}\frac{2\left(-1\right)^j}{n\left(j+2\right)}.$$

Then, I was asked to prove that the series $\sum_{j=0}^\infty c_j$ converges.

Note that (I think) they refer to this series being $\sum_{j=0}^\infty c_j = \sum_{j=0}^\infty\left[\sum_{n=1}^{j+1}\frac{2\cdot\left(-1\right)^j}{n\left(j+2\right)}\right]$

However, when I input

sum [sum (2(-1)^j)/(n(j+2)), n = 1 to j+1], j = 0 to infinity

into Wolfram Alpha, it shows me the right series but it concludes it diverges by the limit test. Is there a mistake I'm making when rewriting? Does the series actually diverge or converge? And if it diverges, how would I prove that using the limit test?

Interestingly, I found that plotting $\sum_{j=0}^x\left(\sum_{n=1}^{j+1}\frac{2\cdot\left(-1\right)^j}{n\left(j+2\right)}\right)$ in Desmos, the graph seems to be alternating but converging.

Answer 1

We are given that the $c_j$ alternate in sign and that $$|c_j|={2 H_{j+1}\over j+2}\to0\qquad(j\to\infty)\ ,$$ where $H_j:=\sum_{k=1}^j{1\over k}\approx \log j$. Furthermore one computes $$|c_j|-|c_{j+1}|={2(j+3)H_{j+1}-2(j+2)H_{j+2}\over(j+2)(j+3)}={2(H_{j+1}-1)\over(j+2)(j+3)}>0\qquad(j\geq1)\ .$$ Altogether this shows that the given series is convergent, by the main theorem on alternating series.

Answer 2

Since $c_j \to 0,$ it's enough to show that the series converges when summed in pairs. So we look at $\sum_{j=1}^{\infty}(c_{2j-1}- c_{2j}).$ Now

$$c_{2j-1}- c_{2j} = -\frac{1}{2j+1}\left(1+\frac{1}{2}+\dots+\frac{1}{2j}\right) + \frac{1}{2j+2}\left(1+\frac{1}{2}+\dots+\frac{1}{2j+1}\right) $$ $$= \left (-\frac{1}{2j+1} + \frac{1}{2j+2}\right)\left(1+\frac{1}{2}+\dots+\frac{1}{2j}\right) + \frac{1}{2j+2}\frac{1}{2j+1}.$$

We're in good shape here. In absolute value, the first term in parentheses is on the order of $1/j^2,$ the second term in parentheses is on the order of $\ln j,$ and the last term is on the order of $1/j^2.$ This shows$\sum_j |c_{2j-1}- c_{2j}|<\infty.$ Thus our series in pairs converges absolutely, hence converges as desired.