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I'm just learning about testing convergence of series of real numbers and I'm stuck with this problem.

Investigate the convergence of the series $$ \sum \limits_{n=1}^\infty \frac{1}{n^2 - \log \, n} $$

My first attempt was to use the "Direct Comparison Test". Looking at the behavior near $ \infty $ (that is, looking at the terms that dominate), we have

$$ \frac{1}{n^2 - \log \, n} \thicksim \frac{1}{n^2} $$

We note that

$$ \frac{1}{n^2 - \log \, n} \ge \frac{1}{n^2} \; \forall n \in \mathbb N $$

The series $ \sum \limits_{n=1}^\infty \frac{1}{n^2} $ converges by the p-series test ($ p \gt 1 $) but I can't conclude anything since the terms of the orginal series are greater than the terms $ \frac{1}{n^2} \forall n \in \mathbb N $.

The "Integral Test" also clearly fails here since there's no obvious substitution for the integrand.

glpsx
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1 Answers1

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$$a_n=\frac2{n^2}$$ $$b_n=\frac1{n^2-\log n}$$

Since $\log n<\frac{n^2}2$,

$$-\log n>-\frac{n^2}2$$

$$n^2-\log n>n^2-\frac{n^2}2$$

$$n^2-\log n>\frac{n^2}2$$

Taking strictly decreasing function $\frac1x$ from both sides: $$b_n=\frac1{n^2-\log n}<\frac2{n^2}<a_n$$

while $a_n$ does converge.

Arashium
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  • This is helpful but I don't understand how you went from $ \log ,n < \frac{n^2}{2} $ to $ n^2 - \log , n > \frac{n^2}{2} $. Could you please explain me this part? – glpsx Oct 25 '15 at 11:50
  • $a<b$ to $-b<-a$ to $c-b<c-a$ or $c-a>c-b$. – Lutz Lehmann Oct 25 '15 at 12:43
  • @VonKar, Updated the answer – Arashium Oct 25 '15 at 13:51
  • The development is very helpful, thanks. – glpsx Oct 25 '15 at 13:53
  • The disadvantage of the Direct Comparison Test is that you can end up in messy inequalities and get bogged down. There is a version that some call the Limit Comparison Test that you must also have learned. It is really just the same test but it relies on your skills with limits instead of your skills with inequalities. If one test works so will the other but with different details. Try both on this problem and see which you prefer. On an exam (probably looming soon) you might find the latter test a bit quicker. – B. S. Thomson Oct 25 '15 at 22:08