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I am having difficulty with the following problem.

Determine if the series $\sum_{n=1}^\infty \sqrt[3]{n^2+4}\cdot[\ln(n^2+2)-\ln(n^2+1)]$ is convergent or divergent.

Here is what I did so far:

Let $a_n=\sqrt[3]{n^2+4}\cdot[\ln(n^2+2)-\ln(n^2+1)]=\sqrt[3]{n^2+4}\cdot\ln\frac{n^2+2}{n^2+1}$, then $\lim_{x\to\infty}a_n=0$. So, the nth Term Test is inconclusive. Also, it seems that I am not supposed to use the Integral Test here.

Then I tried to use the Direct Comparison test but I could not come up with a second series whose terms are bigger than $a_n$'s and convergent or whose terms are smaller than $a_n$'s and divergent. Same with the Limit Comparison Test, I could not find a second sequence $b_n$ so that I can reach a conclusion. Any help will be appreciated.

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Perhaps with some more direct calculus, and for $\;x\ge 0\;$ : define

$$f(x):=\log(1+x)-x\;,\;\;\text{so}\;\;f'(x)=\frac1{1+x}-1=-\frac x{1+x}<0\implies f\searrow$$

monotone descending, and thus

$$\forall\;x>0\;,\;\;f(x)< f(0)=0\implies \log(1+x)<x\;\;,\;\;\;\forall\;x>0$$

So now apply the comparison test (of course, the series is positive):

$$\sqrt[3]{n^2+4}\,\,\log\left(1+\frac1{n^2+1}\right)\le\sqrt[3]2\cdot n^{2/3}\cdot\frac1{n^2+1}\le\sqrt[3]2\,\,n^{-4/3}$$

and since $\;\sum\limits_{n=1} n^p\;$ converges whenever $\;p<-1\;$, we get that our series converges.

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$\log (n^2 + 2)/(n^2+1) = \log (1+2/n^2)/(1+1/n^2) \sim 1/n^2.$ So, $a_n\sim n^{-4/3},$ so your sum is convergent.

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  • $\begingroup$ By which test?-- $\endgroup$ – frosh Feb 16 '16 at 22:08
  • $\begingroup$ @BobSacamano Well, $\sum_{n=2^k}^{2^{k+1}} 1/n^a \leq 2^k/2^{k a} = 1/2^{k(a-1)}.$ So, your series converges by comparison to the geometric series. $\endgroup$ – Igor Rivin Feb 16 '16 at 22:12
  • $\begingroup$ I see, but so far I was not aware that we can use $a_n$'s approximation to make a decision on the convergence of $\sum a_n$. What is the reason behind? $\endgroup$ – frosh Feb 16 '16 at 22:16
  • $\begingroup$ @BobSacamano Because approximation really means that you are within a constant factor of the approximating quantity (eventually), so you can use the comparison test. $\endgroup$ – Igor Rivin Feb 16 '16 at 22:26
  • $\begingroup$ Okay, last but not least, how did you find that $\log (1+2/n^2)/(1+1/n^2) \sim 1/n^2$? $\endgroup$ – frosh Feb 16 '16 at 22:36

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