For $$\sum_{n=1}^{\infty}\ln(\frac{1}{n^2})$$ How do you tell if this series converges or diverges and which test do you use? I tried comparison test but it converges for me when it's supposed to diverge.
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What did you compare with? – enedil Mar 18 '19 at 23:04
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I compared it with ln(1/n) – Random Student Mar 18 '19 at 23:04
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1When you have a series $\sum_{n=1}^\infty a_n$ the first thing you do is to see whether $a_n\to 0$ – clark Mar 18 '19 at 23:09
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Alright so in this case, it moves towards negative infinity. What if it did go to zero? – Random Student Mar 18 '19 at 23:11
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But $\ln(n^{-2})$ doesn't go to zero, so asking the hypothetical "what if it did" is pointless. If you are asking about the general case, there are far too many techniques to say in a comment and it depends heavily on what the actual series is. – JMoravitz Mar 18 '19 at 23:12
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Use the divergence test stating that a series is guaranteed to diverge if the series terms don’t go to zero in the limit.
If $\lim_{n\rightarrow \infty} a_n\neq 0\Rightarrow \sum_n^\infty a_n$ diverges.
$\lim_{n\rightarrow \infty}ln(\frac{1}{n^2})=-\infty\neq 0 \Rightarrow $ $\sum_{n=1}^{\infty}ln(\frac{1}{n^2})$ diverges.
Zack King
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2No problem! If the series terms do happen to go to zero the series may or may not converge! This convergence test can only prove that a series diverges, not that it will converge. You should look up the harmonic series as an example – Zack King Mar 18 '19 at 23:28
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$$\ln \left( \frac{1}{n^2} \right) = - 2 \ln(n)$$
Therefore the general term of your series does not tend to $0$, so your series diverges.
TheSilverDoe
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