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

Sep 7, 2014

If you are trying determine the conergence of $\sum \left\{{a}_{n}\right\}$, then you can compare with $\sum {b}_{n}$ whose convergence is known.

If $0 \le q {a}_{n} \le q {b}_{n}$ and $\sum {b}_{n}$ converges, then $\sum {a}_{n}$ also converges.
If ${a}_{n} \ge q {b}_{n} \ge q 0$ and $\sum {b}_{n}$ diverges, then $\sum {a}_{n}$ 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: ${\sum}_{n = 1}^{\infty} \frac{\sin n + 1}{n}$
Since $\frac{\sin n + 1}{n} \ge q \frac{1}{n}$ and $\setminus {\sum}_{n = 1}^{\infty} \frac{1}{n}$ is the harmonic series, which is divergent, we may conclude that ${\sum}_{n = 1}^{\infty} \frac{\sin n + 1}{n}$ is also divergent by the direct comparison test.

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