The Ratio Test is a very nice tool to use for determining whether or not a series converges, but it can get a bit complicated. Here it is:
If we have a series \(\sum a_n\), let there be a number \(L = \lim_{n\to\infty} \left|\dfrac{a_{n+1}}{a_n}\right|\).
– If \(L < 1\), the series converges absolutely. – If \(L = 1\) or \(L\) does not exist, the test fails and we will need a different test. – If \(L > 1\), the series diverges.
Sounds simple, right? It can be surprisingly hard sometimes to find \(L\), and often you need to get a little creative with it. Let’s prove that this test works first though.
Proof of the Ratio Test
We will treat the three cases, \(L < 1\), \(L = 1\), and \(L > 1\) separately. Let’s start with the easy part:
– If \(L > 1\), that means that \(\lim_{n\to\infty} a_n\) diverges, so the series must also diverge. For example, for the series \(\sum_{n=0}^\infty \dfrac{5}{3^n}\), \(L = \dfrac{5}{3}\), and the terms of the series clearly diverge to infinity, meaning that the series diverges.
– Now let’s consider if \(L < 1\). In that case, there will be some number \(r\) such that \(L < r < 1\) and, for some value of a number \(N\) high enough, for all \(n \ge N\),
\[
\left|\dfrac{a_{n+1}}{a_n}\right| < r \Rightarrow \left|a_{n+1}\right| < \left|a_n\right| \cdot r
\]
We can then compare this series to a geometric series with common ratio \(r\), because, for every \(n \ge N\), \(a_n \le a_N \cdot r^{n-N}\). We know that a geometric series with a common ratio less than 1 converges, so, because \(r < 1\), \(\sum a_n\) must as well.
– Now for when \(L = 1\). For the proof that this test is inconclusive, we just need to provide an example of a convergent series where \(L = 1\) and a divergent series where \(L = 1\). Of course, there are infinitely many convergent and divergent series where \(L = 1\), but here are some simple examples:
Convergent series where \(L = 1\): \(\sum_{n=1}^\infty \dfrac{1}{n^2}\). \(\lim_{n\to\infty} \dfrac{n^2}{(n+1)^2} = L = 1\) by the Dominant Term Test, so this series does satisfy the conditions that this series converges (by the p-series test) and \(L = 1\).
Divergent series where \(L = 1\): \(\sum_{n=1}^\infty n\). That’s right, one of the simplest series there is. \(\lim_{n\to\infty} \dfrac{n+1}{n} = L = 1\), so this does satisfy the conditions that \(L = 1\) and the series diverges.
And if \(L\) does not exist, we can do the same thing as \(L = 1\) by showing a convergent and divergent series where \(L\) does not exist.
Convergent series: \(\sum_{n=1}^\infty \dfrac{(-1)^n + 1}{n^2}\). As we can see, the series is 0 when \(n\) is odd, so \(L\) does not exist as \(\dfrac{a_{n+1}}{a_n}\) is either 0 or undefined and the limit as \(n \to \infty\) does not converge to anything. This series converges because, if we remove all the zero terms, we get the series \(\sum_{n=1}^\infty \dfrac{1}{2n^2}\), which converges by comparison to \(\sum_{n=1}^\infty \dfrac{1}{n^2}\).
Divergent series: \(\sum_{n=1}^\infty \sin n\). In this case, \(L\) does not exist because \(\sin n\) is an oscillating function, so \(L\) never settles down to any specific value. In the same way, the series diverges because the sequence \(\sin n\) also diverges, because \(\lim_{x\to\infty} \sin x\) does not exist.
Conclusion
You now should know a lot more about the Ratio Test, and, maybe most importantly, when it fails. The Ratio Test is a powerful weapon to add to your arsenal to help you through one of the most complex calculus topics: determining whether a series is convergent or divergent. Good luck on your future math journeys!
