20. Convergence of Positive Series

c.1. The Simple Comparison Test

To apply the test, given a series \(\displaystyle \sum_{n=n_o}^\infty a_n\) we try to find either a convergent series \(\displaystyle \sum_{n=n_o}^\infty b_n\) which is bigger than \(\displaystyle \sum_{n=n_o}^\infty a_n\) or a divergent series \(\displaystyle \sum_{n=n_o}^\infty b_n\) which is smaller than \(\displaystyle \sum_{n=n_o}^\infty a_n\). To construct the terms \(b_n\), we look at the terms \(a_n\) and throw away pieces which are negligible for large \(n\). Then we must check the convergence of \(\displaystyle \sum_{n=n_o}^\infty b_n\) and the inequality between \(b_n\) and \(a_n\). Hopefully, this will become clearer in the examples:

Determine if \(\displaystyle \sum_{n=1}^\infty \dfrac{n^2-\sin^2(n)}{n^2+n^4}\) is convergent or divergent.

We start with \(a_n=\dfrac{n^2-\sin^2(n)}{n^2+n^4}\).
For large \(n\), the term \(\sin^2(n)\) is negligible compared to \(n^2\).   Why?

Also the term \(n^2\) is negligible compared to \(n^4\).   Why?

So we take \(b_n=\dfrac{n^2}{n^4}=\dfrac{1}{n^2}\). Now we need to compare \(b_n\) to \(a_n\).

First, \(n^2+n^4 \gt n^4\), because there is more on the left. So \(\dfrac{1}{n^2+n^4} \lt \dfrac{1}{n^4}\).   Why?

Second, \(n^2-\sin^2(n) \lt n^2\), because there is less on the left. Multiplying these two inequalities, we have \[ \dfrac{n^2-\sin^2(n)}{n^2+n^4} \lt \dfrac{n^2}{n^4}=\dfrac{1}{n^2} \qquad \text{or} \qquad a_n \le b_n. \] Finally, \(\displaystyle \sum_{n=1}^\infty b_n =\sum_{n=1}^\infty \dfrac{1}{n^2}\) converges because it is a \(p\)-series with \(p=2>1\). So \(\displaystyle \sum_{n=1}^\infty a_n =\sum_{n=1}^\infty \dfrac{n^2-\sin^2(n)}{n^2+n^4}\) also converges.

Determine if \(\displaystyle \sum_{n=1}^\infty \dfrac{e^n}{ne^n-e^{-n}}\) is convergent or divergent.

We have \(a_n=\dfrac{e^n}{ne^n-e^{-n}}\). For large \(n\), the term \(e^{-n}\) is negligible compared to \(ne^n\). So we take \(b_n=\dfrac{e^n}{ne^n}=\dfrac{1}{n}\). Now \(\displaystyle \sum_{n=1}^\infty b_n=\sum_{n=1}^\infty \dfrac{1}{n}\) diverges because it is the harmonic series. Finally, we need to compare \(a_n\) and \(b_n\). First, \(ne^n-e^{-n} \lt ne^n\) because there is less on the left. So \(\dfrac{1}{ne^n-e^{-n}} \gt \dfrac{1}{ne^n}\), and consequently, \(\dfrac{e^n}{ne^n-e^{-n}} \gt \dfrac{e^n}{ne^n}=\dfrac{1}{n}\) or \(a_n \gt b_n\). Therefore, \(\displaystyle \sum_{n=1}^\infty a_n\) also diverges.