How do you use the direct comparison test for infinite series?

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How do you use the direct comparison test for infinite series?

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Answer 1 · 2014-09-07T11:29:05

If you are trying determine the conergence of $\sum {a_{n}}$ $sum{a_n}$ , then you can compare with $\sum b_{n}$ $sum b_n$ whose convergence is known.

If $0 \leq a_{n} \leq b_{n}$ $0 leq a_n leq b_n$ and $\sum b_{n}$ $sum b_n$ converges, then $\sum a_{n}$ $sum a_n$ also converges.
If $a_{n} \geq b_{n} \geq 0$ $a_n geq b_n geq 0$ and $\sum b_{n}$ $sum b_n$ diverges, then $\sum a_{n}$ $sum a_n$ also diverges.

This test is very intuitive since all it is saying is that if the larger series comverges, then the smaller series also converges, and if the smaller series diverges, then the larger series diverges.

Let us look at some examples.

Example 1: $\infty \sum n = 1 \frac{sin n + 1}{n}$ $sum_{n=1}^{infty}{sin n+1}/{n}$
Since $\frac{sin n + 1}{n} \geq \frac{1}{n}$ ${sin n+1}/n geq 1/n$ and $\infty \sum n = 1 \frac{1}{n}$ $\sum_{n=1}^{infty}1/n$ is the harmonic series, which is divergent, we may conclude that $\infty \sum n = 1 \frac{sin n + 1}{n}$ $sum_{n=1}^{infty}{sin n+1}/{n}$ is also divergent by the direct comparison test.

Example 2: $\infty \sum n = 1 \frac{{cos}^{2} n}{n^{\frac{3}{2}}}$ $sum_{n=1}^{infty}{cos^2n}/n^{3/2}$
Since $\frac{{cos}^{2} n}{n^{\frac{3}{2}}} \leq \frac{1}{n^{\frac{3}{2}}}$ ${cos^2n}/{n^{3/2}} leq 1/n^{3/2}$ and $\infty \sum n = 1 \frac{1}{n^{\frac{3}{2}}}$ $sum_{n=1}^{infty}1/n^{3/2}$ is a convergent p-series (p>1), we may conclude that $\infty \sum n = 1 \frac{{cos}^{2} n}{n^{\frac{3}{2}}}$ $sum_{n=1}^{infty}{cos^2n}/n^{3/2}$ is also convergent by the dirct comparison test.

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How do you use the direct comparison test for infinite series?

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