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Limit comparison test

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Title: Limit comparison test  
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Subject: Convergence tests, Mean value theorem, Series (mathematics), Calculus, Comparison test
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Limit comparison test

In mathematics, the limit comparison test (LCT) (in contrast with the related direct comparison test) is a method of testing for the convergence of an infinite series.


Suppose that we have two series \Sigma_n a_n and \Sigma_n b_n with a_n, b_n \geq 0 for all n.

Then if \lim_{n \to \infty} \frac{a_n}{b_n} = c with 0 < c < \infty then either both series converge or both series diverge.


Because \lim \frac{a_n}{b_n} = c we know that for all \varepsilon there is an integer n_0 such that for all n \geq n_0 we have that \left| \frac{a_n}{b_n} - c \right| < \varepsilon , or what is the same

- \varepsilon < \frac{a_n}{b_n} - c < \varepsilon
c - \varepsilon < \frac{a_n}{b_n} < c + \varepsilon
(c - \varepsilon)b_n < a_n < (c + \varepsilon)b_n

As c > 0 we can choose \varepsilon to be sufficiently small such that c-\varepsilon is positive. So b_n < \frac{1}{c-\varepsilon} a_n and by the direct comparison test, if a_n converges then so does b_n .

Similarly a_n < (c + \varepsilon)b_n , so if b_n converges, again by the direct comparison test, so does a_n .

That is both series converge or both series diverge.


We want to determine if the series \sum_{n=1}^{\infty} \frac{1}{n^2 + 2n} converges. For this we compare with the convergent series \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} .

As \lim_{n \to \infty} \frac{1}{n^2 + 2n} \frac{n^2}{1} = 1 > 0 we have that the original series also converges.

See also

External links

  • Pauls Online Notes on Comparison Test
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