In this lecture we will see a few basic tests that can help us to determine if a series converge or diverge without explicitly computing the limits of the partial sums.
Tianran Chen
Department of Mathematics
Auburn University at Montgomery
We learned how to determine if a series converges by computing limits of partial sums. Unfortunately, such direct computation can be quite difficult in many cases. In those case, we often prefer to use indirect convergence/divergence tests which we will outline in the following.
If $\lim_{k \to \infty} a_k \ne 0$ (or does not exist), then $\sum_{k=1}^\infty a_k$ diverges.
This is a fairly straightforward consequence of how we defined convergence/divergence in terms of the limit of the partial sums.
It is super important to understand that the converse of the above statement is not true: Having $\lim_{k \to \infty} a_k = 0$ does not tell us that $\sum_{k=1}^\infty a_k$ converges (e.g., the Harmonic series $\sum_{k=1}^\infty 1 / k$).
Problem. Determine if $\sum_{k=1}^\infty \cos( 1 / k )$ converges/diverges.
Solution. We can see that, since $\cos$ is a continuous function, \[ \lim_{k \to \infty} \cos( 1 / k ) = \cos \left( \lim_{k \to \infty} 1 / k \right) = \cos (0) = 1. \] Therefore, by the divergence test, we can conclude that this series diverge.
What about the series $\sum_{k=1}^\infty \sin( 1 / k )$ ?
This is an example in which the divergence test tells us nothing.
It is important to understand that even though \[ \lim_{k \to \infty} \sin \left( \frac{1}{k} \right) = \sin \left( \lim_{k \to \infty} \frac{1}{k} \right) = \sin (0) = 0, \] we cannot conclude that the series is therefore convergent (the divergent test can help us to confirm a series is divergent, it cannot, however, tell us if a series is convergent.
In other words, having the limit of the terms be zero is a necessary condition for convergence, but that is not an sufficient condition.
This test turns the problem of determining the convergence property of a series into the problem of determining the convergence property of an improper integral for which we have more advanced tools.
If $\sum_{k=1}^\infty a_k$ is a series and $f$ is a continuous and decreasing function such that $f(k) = a_k > 0$ for any positive integer $k$, then \[ \sum_{k=1}^\infty a_k \quad\text{and}\quad \int_1^\infty f(x) dx \quad\text{both converge or both diverge}. \]
It is important to understand that the integral test is tests for convergence. In general, it cannot tell you which value a series converges to.
Problem. Determine if $\sum_{k=1}^\infty \frac{3}{\sqrt{2k + 4}}$ is convergent or divergent.
Solution. To determine if this series is convergent using integral test, we can compute the improper integral \[ \int_1^\infty \frac{3}{ \sqrt{2x + 4} } dx = \lim_{t \to \infty} \int_1^t \frac{3}{ \sqrt{2x + 4} } dx = \lim_{t \to \infty} \left. 3 \sqrt{2x+4} \, \right|_1^t \] which diverges (goes to $\infty$). By the integral test, we can conclude that the series $\sum_{k=1}^\infty \frac{3}{\sqrt{2k + 4}}$ is also divergent.
Problem. For a positive real number $p \ne 1$, determine if the series $\sum_{k=1}^\infty 1 / k^p$ is convergent or divergent.
Solution. We can compute the improper integral $\int_1^\infty \frac{1}{ x^p } dx$, which turns into \[ \lim_{t \to \infty} \int_1^\infty \frac{1}{ x^p } dx = \lim_{t \to \infty} \left. \frac{x^{1-p}}{ (1-p) } \right|_1^t = \lim_{t \to \infty} \frac{t^{1-p}}{ (1-p) } - \frac{1}{1-p} \] which converge if $p > 1$ and diverge if $p < 1$. By the integral test, \[ \sum_{k=1}^\infty \frac{1}{k^p} \quad \begin{cases} \text{converge if} & p > 1 \\ \text{diverge if} & p < 1 \end{cases} \]
The solution to the last problem is particularly useful and thus deserves its own test. Series of the form $\sum_{k=1}^\infty 1 / k^p$ for some positive real number $p$ appear very frequently and earned the nickname $p$-series.
\[ \sum_{k=1}^\infty \frac{1}{k^p} \quad \begin{cases} \text{converges if }p > 1 \\ \text{diverges if }p \le 1. \end{cases} \]
Note that the special case of $p=1$ is exactly the harmonic series, which we know is divergent.
The tests stated above are about series of the form $\sum_{k=1}^\infty a_k$. However, it is important to understand that the starting index does not matter. Indeed, one could ignore a finite number of terms.
For any positive integer $N$, the two series \begin{align*} &\sum_{k=1}^\infty a_k &&\text{and} & &\sum_{k=N}^\infty a_k &&\text{both converge or both diverge}. \end{align*}