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		<title>Morrie&#039;s law</title>
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		<summary type="html">&lt;p&gt;201.141.177.206: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Calculus|Series}}&lt;br /&gt;
&lt;br /&gt;
In [[mathematics]], the &#039;&#039;&#039;limit comparison test (LCT)&#039;&#039;&#039; (in contrast with the related [[direct comparison test]]) is a method of testing for the convergence of an [[infinite series]].&lt;br /&gt;
&lt;br /&gt;
== Statement ==&lt;br /&gt;
&lt;br /&gt;
Suppose that we have two series &amp;lt;math&amp;gt; \Sigma_n a_n &amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\Sigma_n b_n&amp;lt;/math&amp;gt; with &amp;lt;math&amp;gt; a_n, b_n \geq 0 &amp;lt;/math&amp;gt; for all &amp;lt;math&amp;gt; n&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Then if &amp;lt;math&amp;gt; \lim_{n \to \infty} \frac{a_n}{b_n} = c&amp;lt;/math&amp;gt; with &amp;lt;math&amp;gt; 0 &amp;lt; c &amp;lt; \infty &amp;lt;/math&amp;gt; then either both series converge or both series diverge.&lt;br /&gt;
&lt;br /&gt;
== Proof ==&lt;br /&gt;
&lt;br /&gt;
Because &amp;lt;math&amp;gt; \lim \frac{a_n}{b_n} = c &amp;lt;/math&amp;gt; we know that for all &amp;lt;math&amp;gt; \varepsilon &amp;lt;/math&amp;gt; there is an integer &amp;lt;math&amp;gt;n_0&amp;lt;/math&amp;gt; such that for all &amp;lt;math&amp;gt;n \geq n_0 &amp;lt;/math&amp;gt; we have that &amp;lt;math&amp;gt; \left| \frac{a_n}{b_n} - c \right| &amp;lt; \varepsilon &amp;lt;/math&amp;gt;, or what is the same&lt;br /&gt;
&lt;br /&gt;
: &amp;lt;math&amp;gt; - \varepsilon &amp;lt; \frac{a_n}{b_n} - c &amp;lt; \varepsilon &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
: &amp;lt;math&amp;gt; c - \varepsilon &amp;lt; \frac{a_n}{b_n} &amp;lt; c + \varepsilon &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
: &amp;lt;math&amp;gt; (c - \varepsilon)b_n &amp;lt; a_n &amp;lt; (c + \varepsilon)b_n &amp;lt;/math&amp;gt;&lt;br /&gt;
As &amp;lt;math&amp;gt; c &amp;gt; 0 &amp;lt;/math&amp;gt; we can choose &amp;lt;math&amp;gt; \varepsilon &amp;lt;/math&amp;gt; to be sufficiently small such that &amp;lt;math&amp;gt; c-\varepsilon &amp;lt;/math&amp;gt; is positive.&lt;br /&gt;
So &amp;lt;math&amp;gt; b_n &amp;lt; \frac{1}{c-\varepsilon} a_n &amp;lt;/math&amp;gt; and by the [[direct comparison test]], if &amp;lt;math&amp;gt;a_n&amp;lt;/math&amp;gt; converges then so does &amp;lt;math&amp;gt; b_n &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Similarly &amp;lt;math&amp;gt; a_n &amp;lt; (c + \varepsilon)b_n &amp;lt;/math&amp;gt;, so if &amp;lt;math&amp;gt; b_n &amp;lt;/math&amp;gt; converges, again by the direct comparison test, so does &amp;lt;math&amp;gt; a_n &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
That is both series converge or both series diverge.&lt;br /&gt;
&lt;br /&gt;
==Example==&lt;br /&gt;
&lt;br /&gt;
We want to determine if the series &amp;lt;math&amp;gt; \sum_{n=1}^{\infty} \frac{1}{n^2 + 2n} &amp;lt;/math&amp;gt; converges.  For this we compare with the convergent series &amp;lt;math&amp;gt; \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} &amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
As &amp;lt;math&amp;gt; \lim_{n \to \infty} \frac{1}{n^2 + 2n} \frac{n^2}{1} = 1 &amp;gt; 0 &amp;lt;/math&amp;gt; we have that the original series also converges.&lt;br /&gt;
&lt;br /&gt;
==See also==&lt;br /&gt;
* [[Convergence tests]]&lt;br /&gt;
* [[Direct comparison test]]&lt;br /&gt;
&lt;br /&gt;
== External links ==&lt;br /&gt;
* [http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx Pauls Online Notes on Comparison Test ]&lt;br /&gt;
&lt;br /&gt;
[[Category:Convergence tests]]&lt;br /&gt;
[[Category:Articles containing proofs]]&lt;/div&gt;</summary>
		<author><name>201.141.177.206</name></author>
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