This tests utilises the ‘bounds’ of other Series, usually geometric or p-harmonic, to test a given series. It’s similar to the Sandwich Theorem, but doesn’t require two other series, since a series can either converge to a value or diverge (approach infinity).
Definition
Let be a sequence of real numbers with positive terms only, and be it’s respective series.
-
If there exists another series, , such that then if diverges, also diverges. I.e. if we have a series that is always smaller than , and that series diverges, then the bigger series, must also diverge.
-
Similarly, if there exists a series, such that then if converges, also converges
It’s best to use geometric or harmonic series because their conditions for converging or diverging are simple.
https://www.desmos.com/calculator/8op6xpfuks
Examples
1: Divergence
**Determine if the following series converges or diverges:
\sum_{n=1}^{\infty} \frac{n^2+4}{n^3+5}$$** \ We can see that, by looking at the fastest growing terms, the series approximately diverges: $\frac{n^2+4}{n^3+5} \sim \frac{n^2}{n^3} = \frac{1}{n}$ which is an example of a [p-harmonic series](Series.md#p-Harmonic%20Series). Now, we need to prove this using inequalities. \ $\frac{n^2+4}{n^3+5} > \frac{n^2}{n^3+5} > \frac{n^2}{n^3 + 5n^3}$ (as $n\ge 0 \ \forall \ n$) $\frac{n^2}{6n^3} = \frac{1}{6n}$ So, $\frac{1}{6n} \leq \frac{n^2+4}{n^3+5}$ for every term of n, meaning we can use the comparison test. \\sum_{n=1}^\infty \frac{1}{6n} \text{diverges, as } \frac{1}{6} \text{ is a constant}(using [p-harmonic series](Series.md#p-Harmonic%20Series), as well as the [properties of divergent series](Series.md#Properties%20Of%20Series)) \ Hence,\sum_{n=1}^{\infty} \frac{n^2+4}{n^3+5} \text{diverges}