Root Test for Series Convergence
Summarize
What is the root test
The root test is a method used to determine whether an infinite series converges or diverges. It is particularly useful when each term of the series involves an expression raised to the $n$-th power, such as exponentials or roots. Suppose we have a series with positive terms of the form:
$$ \sum_{n=1}^{+\infty} a_n $$
Assume that the limit exists and is finite:
$$ \limsup_{n \to \infty} \sqrt[n]{|a_n|} = L $$
We use $\limsup$ because it is always defined for real sequences that are bounded from below. This makes the root test reliable even when the usual limit $\lim_{n \to \infty} \sqrt[n]{|a_n|}$ does not exist. By taking the limit superior, the test can still detect the long-term behavior of the sequence and determine whether the series converges or diverges.
If the limit $L$ exists, three cases can occur:
- If $L < 1$, the series converges absolutely.
- If $L > 1$ or $L = \infty$, the series diverges.
- If $L = 1$, the test is inconclusive.
We use the absolute value $|a_n|$ to apply the root test to the absolute convergence of the series. This ensures the test works even if the terms $a_n$ are negative or alternate in sign, allowing us to focus only on their magnitude.
How to recognize when to apply the root test
The root test is especially useful when the general term of a series is expressed in the form $a_n = (b_n)^n$, that is, when the entire term is raised to the power of $n$. In such cases, taking the $n$-th root of $a_n$ simplifies the expression significantly and often leads directly to a manageable limit.
By contrast, other tests may become more complicated in this context, especially when the ratio $\frac{a_{n+1}}{a_n}$ does not simplify easily or when factorials or exponential terms are involved in a way that makes limits more difficult to compute.
The root test is particularly effective in the following situations:
- When $a_n = (f(n))^n$, with $f(n) > 0$
- When the terms involve exponential-like growth or decay
- When the limit $\sqrt[n]{|a_n|}$ is easier to compute than $\frac{a_{n+1}}{a_n}$
In all other cases, if the general term is not raised to the $n$-th power other tests may be more suitable.
Proof
Consider the case of absolute convergence, with $L < 1$. There exists a real number $r$ such that:
$$ L < r < 1 $$
By the definition of $\limsup$, there exists an integer $N$ such that for all $n \geq N$:
$$ \sqrt[n]{|a_n|} < r \quad \rightarrow \quad |a_n| < r^n $$
So the tail of the series satisfies:
$$ \sum_{n=N}^{\infty} |a_n| < \sum_{n=N}^{\infty} r^n $$
Since $0 < r < 1$, the geometric series $\sum r^n$ converges. Therefore, by the comparison test, the tail $\sum_{n=N}^{\infty} |a_n|$ converges, and so does the entire series $\sum |a_n|$.
Consider the case in which the series diverges, with $L > 1$. By definition of $\limsup$, for infinitely many indices $n$, we have:
$$ \sqrt[n]{|a_n|} > r \quad \text{for some } r > 1 $$
Thus, $|a_n| > r^n$ infinitely often. But since $r^n \to \infty$, we know that $|a_n| \nrightarrow 0$.
The necessary condition for convergence of $\sum a_n$ fails, so the series diverges.
In the final case nothing can be concluded, namely when $L = 1$. For every $\varepsilon > 0$ we can find infinitely many $n$ such that:
$$ \sqrt[n]{|a_n|} > 1 - \varepsilon \quad \text{and} \quad \sqrt[n]{|a_n|} < 1 + \varepsilon $$
This range includes both convergent and divergent behaviors. For example, the harmonic series $a_n = \frac{1}{n}$ has $\sqrt[n]{|a_n|} \to 1$ and diverges. The p-series $a_n = \frac{1}{n^2}$ also has $\sqrt[n]{|a_n|} \to 1$, but it converges. So, the test is inconclusive when $L = 1$.
Example
Consider the series:
$$ \sum_{n=1}^{\infty} \left( \frac{3n}{5n + 2} \right)^n $$
We determine whether this series converges or diverges. Each term has the form $a_n = (\text{expression})^n$, so the root test applies and is simpler than other methods.
We define the sequence:
$$ a_n = \left( \frac{3n}{5n + 2} \right)^n $$
To apply the root test, we evaluate the limit superior of the $n$-th root of $a_n$:
$$ \limsup_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \left( \frac{3n}{5n + 2} \right) $$
Since the terms are positive, we omit the absolute value and simplify:
$$ \frac{3n}{5n + 2} = \frac{3}{5 + \frac{2}{n}} \longrightarrow \frac{3}{5} \quad \text{as } n \to \infty $$
Therefore, we find:
$$ \limsup_{n \to \infty} \sqrt[n]{a_n} = \frac{3}{5} < 1 $$
Since the limit is less than $1$, the root test tells us that the series converges absolutely.
A trigonometric factor raised to the $n$-th power leads to the same kind of computation. We study the series:
$$ \sum_{n=1}^{\infty} \left( \frac{n}{3} \tan \frac{1}{n} \right)^n $$
The terms are positive for every $n \geq 1$, and the whole expression is raised to the power $n$, so the root test is the natural choice. Taking the $n$-th root removes the outer exponent:
$$ \sqrt[n]{a_n} = \frac{n}{3} \tan \frac{1}{n} $$
As $n \to \infty$ the argument $\frac{1}{n}$ tends to zero, where $\tan \frac{1}{n} \sim \frac{1}{n}$ is a remarkable limit. The limit of the root is therefore:
$$ \lim_{n \to \infty} \frac{n}{3} \tan \frac{1}{n} = \lim_{n \to \infty} \frac{n}{3} \cdot \frac{1}{n} = \frac{1}{3} $$
Since the limit equals $\frac{1}{3} < 1$, the series converges.