The Limit Comparison Test

The Limit Comparison Test is a good test to try when a basic comparison does not work (as in Example 3 on the previous slide).  The idea of this test is that if the limit of a ratio of sequences is 0, then the denominator grew much faster than the numerator.  If the limit is infinity, the numerator grew much faster.  If your limit is non-zero and finite, the sequences behave similarly so their series will behave similarly as well.

Limit Comparison Test:  Let $\displaystyle{\sum_{n=1}^\infty a_n}$ and $\displaystyle{\sum_{n=1}^\infty b_n}$ be positive-termed series.  If $$\displaystyle{\lim_{n \to \infty} \frac{a_n}{b_n}}=c,$$ where $c$ is finite, and $c>0$, then either both series converge or both diverge. 


Example 1 (from previous page):  We were trying to determine whether $\displaystyle\sum_{n=1}^\infty\frac{1}{5n+10}$ converges or diverges, and the basic comparison test was not helpful.  DO:  Try the limit comparison test on this series, comparing it to the harmonic series, before reading further.


Solution 1:  It does not matter which series you choose to have terms $a_n$ and $b_n$. 
$\displaystyle\frac{a_n}{b_n}=\frac{\frac{1}{5n+10}}{\frac1n}=\frac{1}{5n+10}\frac{n}{1}=\frac{n}{5n+10}$.  Then $\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{n}{5n+10}=\frac15$ and $0<\frac15<\infty$, so the series behave the same.  Since the harmonic series diverges, so does our series.  DO:  What if we had chosen $a_n$ and $b_n$ the other way around?  Can you see why it doesn't matter?

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Example 2:   To determine whether the series $\displaystyle\sum_{n=1}^\infty \frac{4^n}{2^n+3^n}$ converges or diverges, we'll look for a series that "behaves like" it when $n$ is large. 

Solution 2:  Since we think $$ \frac{4^n}{2^n+3^n}\approx \frac{4^n}{3^n}, $$ when $n$ is large, we'll use $\displaystyle\sum_{n=1}^\infty\left(\frac43\right)^n$ for comparison.  (DO:  Why can we not use the basic comparison test with this series?)  Since $$ \lim_{n\to\infty}\frac{ \frac{4^n}{2^n+3^n}}{\frac{4^n}{3^n}}= \lim_{n\to\infty}\frac{ 3^n}{2^n+3^n}= \lim_{n\to\infty}\frac{ 3^n}{2^n+3^n}\frac{\frac1{3^n}}{\frac1{3^n}}= \lim_{n\to\infty}\frac{1}{\left(\frac{2^n}{3^n}\right)+1}= 1,$$ and $0<1<\infty$, our series are comparable.  Since the geometric series $\displaystyle\sum_{n=1}^\infty\left(\frac43\right)^n$ diverges ($r=\frac43>1$), we can conclude that our original series diverges as well.




The following test, which was discussed in the video, is not explicitly used by most instructors, and is not in most calculus texts -- discuss it with your instructor before using.
Theorem: Suppose that $\displaystyle{\sum_{n=1}^\infty a_n}$ and $\displaystyle{\sum_{n=1}^\infty b_n}$ are positive-termed series, $c$ is a positive constant, and that $N$ is some positive integer.
  • If $\displaystyle{\sum_{n=1}^\infty b_n}$ converges and $a_n \le c b_n$ for all $n>N$, then $\displaystyle{\sum_{n=1}^\infty a_n}$ converges.

  • If $\displaystyle{\sum_{n=1}^\infty b_n}$ diverges and $a_n \ge c b_n$ for all $n>N$, then $\displaystyle{\sum_{n=1}^\infty a_n}$ diverges.