Does the Comparison Test Determine Convergence or Divergence of Series?

In summary, the comparison test is used to determine if a series converges or diverges. In this case, we compared the given series $\sum_{n=1}^\infty \frac{1}{n^2\ln(n)-10}$ with the convergent series $\sum_{n=1}^\infty \frac{1}{n^2}$ and found that for $n \ge 5$, the given series is smaller and therefore converges.
  • #1
karush
Gold Member
MHB
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Use the comparison test to determine if the series series convergences or divergences
$$S_{6}=\sum_{n=1}^{\infty} \dfrac{1}{n^2 \ln{n} -10}$$
ok if i follow the example given the next step alegedly would be...
$$\dfrac{1}{n^2 \ln{n} -10}<\dfrac{1}{n^2 \ln{n}}$$
$\tiny{242 UHM}$
 
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  • #2
note ...

$n^2\ln(n)-10 < n^2\ln(n) \implies \dfrac{1}{n^2\ln(n) - 10} > \dfrac{1}{n^2\ln(n)}$

$\displaystyle \sum \dfrac{1}{n^2\ln(n)} < \sum \dfrac{1}{n^2}$ converges ... so the comparison above isn't going to fly.
 
  • #3
oh ..

why put the $\sum $ back in ??
 
  • #4
karush said:
oh ..

why put the $\sum $ back in ??

The goal is to compare series, not just the nth term.Let's compare \(\displaystyle \sum \frac{1}{n^2\ln(n)-10}\) with the known convergent series \(\displaystyle \sum \frac{1}{n^2}\)

if \(\displaystyle \sum \frac{1}{n^2\ln(n)-10} < \sum \frac{1}{n^2}\) for large enough $n$, then we can say \(\displaystyle \sum \frac{1}{n^2\ln(n)-10}\) converges

first off, we need to find what values of $n$ make \(\displaystyle \frac{1}{n^2\ln(n)-10} < \frac{1}{n^2}\)

for that inequality to be true ...

$n^2\ln(n) - 10 > n^2$

$n^2\ln(n) - n^2 > 10$

$n^2[\ln(n)-1] > 10 \implies n \ge 5$

so, \(\displaystyle \sum_{n=5}^\infty \frac{1}{n^2\ln(n)-10} < \sum_{n=5}^\infty \frac{1}{n^2} \implies \sum_{n=5}^\infty \frac{1}{n^2\ln(n)-10}\) converges.
 

1. What is the Comparison Test on series?

The Comparison Test is a method used to determine whether an infinite series converges or diverges by comparing it to another series whose convergence or divergence is already known.

2. How does the Comparison Test work?

The Comparison Test states that if the terms of a series are always less than or equal to the terms of a convergent series, then the original series must also converge. Similarly, if the terms of a series are always greater than or equal to the terms of a divergent series, then the original series must also diverge.

3. What is the difference between the Comparison Test and the Limit Comparison Test?

The Limit Comparison Test is a variation of the Comparison Test that uses the limit of the ratio of the terms of two series to determine convergence or divergence. It is often easier to use when the terms of the series are more complicated.

4. Can the Comparison Test be used to determine the exact value of a series?

No, the Comparison Test only tells us whether a series converges or diverges. It does not provide the exact value of the series.

5. Are there any limitations to using the Comparison Test?

Yes, the Comparison Test can only be used on series with positive terms. It also requires the comparison series to be known and have a known convergence or divergence. In some cases, the Comparison Test may not be conclusive and other methods may need to be used.

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