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.

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    $\begingroup$ Not sure I am following the debate here... Roughly speaking, $|c_j|$ behaves like $1/(j\log j)$, which ultimately decreases to $0$, hence, if indeed $(|c_j|)$ decreases, then the series $\sum c_j$ obviously converges, right? $\endgroup$ – Did Feb 8 '18 at 14:48
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    $\begingroup$ Marc: en.wikipedia.org/wiki/… $\endgroup$ – Did Feb 8 '18 at 15:06
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Jyrki Lahtonen Feb 11 '18 at 5:00

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.

  • $\begingroup$ How do you prove this last inequality I have been searching that $\endgroup$ – Guy Fsone Feb 8 '18 at 16:29
  • $\begingroup$ The displayed formula is the proof. Note that $H_j:=\sum_{k=1}^j{1\over k}$. $\endgroup$ – Christian Blatter Feb 8 '18 at 16:46
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    $\begingroup$ I am afraid that this answer will only be understandable by those who don't need it. In particular, you need to know that $H_j$ are the harmonic numbers (en.wikipedia.org/wiki/Harmonic_number), how they behave as $j \to \infty$, and what the alternating series test says (en.wikipedia.org/wiki/Alternating_series). The OP probably doesn't know much about this, or anyhow needs help in putting the pieces together. $\endgroup$ – Ant Feb 8 '18 at 18:48

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.


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