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I was supposed to use the convergence tests to determine whether the following series converges, converges absolutely, or diverges.

Almost every test I used was inconclusive, the series is : $$\sum_{k=1}^{\infty} (-1)^k \frac{k\ln(k)}{(k+1)^3}.$$

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  • $\begingroup$ For convergence/divergence, use the alternating series test. For absolute convergence, do a comparison test with $\frac{1}{k^{1.5}}$. $\endgroup$ – angryavian Apr 3 '17 at 17:33
  • $\begingroup$ Have you used the Leibniz criterion? This concludes the convergence of the series. $\endgroup$ – Guillemus Callelus Apr 3 '17 at 17:43
  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. $\endgroup$ – Clement C. Apr 13 '17 at 3:01
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You have that $$ \frac{n\ln n}{(n+1)^3} \sim_{n\to \infty} \frac{\ln n}{n^2} $$ so that, since the (positive) series $\sum_n \frac{\ln n}{n^2}$ converges (e.g., by comparison with a $p$-series $\sum_n \frac{1}{n^{3/2}}$) by theorems of comparison the series $\sum_n (-1)^n \frac{n\ln n}{(n+1)^3}$ converges absolutely.

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Consider,

$$f(x)=\frac{x \ln x}{(x+1)^3}$$

Then,

$$f'(x)=\frac{x-2x\ln x+\ln x+1}{(x+1)^4}$$

One can show that for $x>3$ that we must have $x-2x\ln x+\ln x+1<0$ by considering the monotonically decreasing function $g(x)=x-2x\ln x+\ln x+1$ defined on $(1,\infty)$. I leave it to you to fill in the gaps. Thus we have for $x>3$ that $f'(x)<0$. Hence it follows that,

$$\frac{n \ln n}{(n+1)^3}$$

Is eventually decreasing, and your series converges by the alternating series test.

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