Limit Comparison Test

Section: 
9.4
Date: 
Wednesday, October 6, 2010 - 14:00
AttachmentSize
fall2010math2300_limit-comparison-test.pdf47.79 KB

Math 2300 Section 005 – Calculus II – Fall 2010

9.4 Limit Comparison Test Examples – October 6, 2010

  1. Determine if the given series converges or diverges using the limit comparison test.

    \sum _{{n=1}}^{\infty}\frac{1}{2^{n}-1}

    Solution: The dominant terms here are 1 in the numerator and 2^{n} in the denominator. So, we compare the series terms

    a_{n}=\frac{1}{2^{n}-1}\qquad\text{and}\qquad b_{n}=\frac{1}{2^{n}}.

    The terms b_{n} form a convergent geometric series. And, we find that

    \lim _{{n\rightarrow\infty}}\frac{a_{n}}{b_{n}}=\lim _{{n\rightarrow\infty}}\frac{\displaystyle\frac{1}{2^{n}-1}}{\displaystyle\frac{1}{2^{n}}}=\lim _{{n\rightarrow\infty}}\frac{2^{n}}{2^{n}-1}=\lim _{{n\rightarrow\infty}}\frac{1}{1-1/2^{n}}=1.

    Since we obtain a finite positive number using the limit comparison test, we know that both of the series either diverge or they both converge. We already knew that the series of b_{n} terms converges, so the series in question must also converge.

  2. Determine if the given series converges or diverges using the limit comparison test.

    \sum _{{n=1}}^{\infty}\frac{2n^{2}+3n}{\sqrt{5+n^{5}}}

    Solution: The dominant term in the numerator is 2n^{2}, while the dominant term in the denominator is \sqrt{n^{5}}. So, we will attempt to compare this series to

    \sum _{{n=1}}^{\infty}b_{n}=\sum _{{n=1}}^{\infty}\frac{2n^{2}}{\sqrt{n^{5}}}=\sum _{{n=0}}^{\infty}\frac{2}{\sqrt{n}}=2\sum _{{n=1}}^{\infty}\frac{1}{n^{{1/2}}}

    which we know diverges.

    \lim _{{n\rightarrow\infty}}\frac{a_{n}}{b_{n}}=\lim _{{n\rightarrow\infty}}\left[\left(\frac{2n^{2}+3n}{\sqrt{5+n^{5}}}\right)\left(\frac{\sqrt{n}}{2}\right)\right]=\lim _{{n\rightarrow\infty}}\frac{2n^{{5/2}}+3n^{{3/2}}}{2\sqrt{5+n^{5}}}=\lim _{{n\rightarrow\infty}}\frac{2+\frac{3}{n}}{2\sqrt{\frac{5}{n^{5}}+1}}=1

    Since the limit is finite and positive, we know that the original series diverges as well.

© 2011 Jason B. Hill. All Rights Reserved.