The Root Test

The root test is simply a way to find the "effective ratio" of a series by taking the $n$th root of the $n$th term. The name is literal: you take a root, the $n$th one. It's good for terms where everything is raised to the $n$th power, like $\left(\frac{n}{3n+1}\right)^n$, because the root just peels that power off. The formula $L = \lim_{n\to\infty} \sqrt[n]{|a_n|}$ has only a few pieces: $|a_n|$ (the size of the $n$th term), $\sqrt[n]{\ \ }$ (undo an $n$th power) and $L$ (what's left in the long run). It looks heavier than it is; bear with it.

Here's where the idea comes from. In a geometric series the $n$th term is $ar^n$. Take its $n$th root and you get $\sqrt[n]{a}\cdot r$, and since $\sqrt[n]{a} \to 1$, what's left is just $r$. So the $n$th root of a term recovers the ratio, without ever comparing neighboring terms. That's the same question the ratio test asks, approached from a different side.

The verdicts are the same as the ratio test: $L < 1$ converges absolutely, $L > 1$ diverges, $L = 1$ says nothing.

Why. If $\sqrt[n]{|a_n|} \le r < 1$ eventually, then raising both sides to the $n$th power gives $|a_n| \le r^n$, and that's a convergent geometric series sitting above yours.

Limits that come up:

• $\sqrt[n]{n} = n^{1/n} \to 1$, and $\sqrt[n]{c} \to 1$ for any $c > 0$.
• $\left(1 + \frac{a}{n}\right)^n \to e^a$.

Ratio or root? If the term is something to the $n$th power, the root test strips the power off. If it has factorials, use the ratio test (roots of factorials are messy).

A common mistake. It's tempting to think the root of $n$ or of a constant just disappears, so $\sqrt[n]{n^5}$ is "gone." It isn't gone, but it does go to 1: $\sqrt[n]{n^5} = (n^{1/n})^5 \to 1$. Treat those pieces as 1 in the limit, not as 0, and not as something that grows.

The root test defines the radius of convergence of any power series (the Cauchy–Hadamard formula), which tells you how far from its center a series approximation can be trusted.