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You CAN Ace Calculus

17calculus > infinite series > root test

### Calculus Main Topics

Single Variable Calculus

Multi-Variable Calculus

### Tools

math tools

general learning tools

Root Test

In our experience, the Root Test is the least used series test to test for convergence or divergence (which is why it appears last in the infinite series table). The reason is that it is used only in very specific cases, whereas the other tests can be used for a broader range of problems. However, it can be used to determine convergence or divergence much easier than using other techniques in these very specific cases.

Root Test

For the series $$\sum{a_n}$$, let $$\displaystyle{L = \lim_{n \to \infty}{\sqrt[n]{|a_n|}}}$$.

Three cases are possible depending on the value of L.
$$L < 1$$: The series converges absolutely.
$$L = 1$$: The Root Test is inconclusive.
$$L> 1$$: The series diverges.

When To Use The Root Test

The Root Test is used when you have a function of n that also contains a power with an n. The idea is to remove or change the n in the power. The test itself is fairly straight-forward. You just need some practice using it to know under what conditions it is best to use it. One good way to learn when to use this test is by building example pages as described in the Study Techniques section.

A powerful rule that will be useful when using the root test is $$\displaystyle{ \lim_{n \to \infty}{\ln(f(n))} = \ln\left( \lim_{n \to \infty}{f(n)} \right) }$$
This is true because the natural log function is continuous. Here are a few additional rules that may be useful.

 $$\displaystyle{ \lim_{n \to \infty}{ \sqrt[n]{p}} = 1 }$$ $$p$$ is a positive constant $$\displaystyle{ \lim_{n \to \infty}{\sqrt[n]{n^p}} = 1 }$$ $$p$$ is a positive constant $$\displaystyle{ \lim_{n \to \infty}{\sqrt[n]{n}} = 1 }$$ special case of previous line with $$p=1$$ $$\displaystyle{ \lim_{n \to \infty}{\sqrt[n]{\ln~n}} = 1 }$$ $$\displaystyle{ \lim_{n \to \infty}{\sqrt[n]{n!}} = \infty }$$

We prove the first four rules above as practice problems on the L'Hôpital's Rule page. The last one can be proven using Stirlings Formula.

### Search 17Calculus

Practice Problems

Instructions - - Unless otherwise instructed, determine whether these series converge or diverge using the root test.

 Level A - Basic

Practice A01

$$\displaystyle{\sum_{n=1}^{\infty}{\left[\frac{n^2+1}{2n^2+1}\right]^n}}$$

solution

Practice A02

$$\displaystyle{\sum_{n=1}^{\infty}{\left[3^n e^{-n}\right]}}$$

solution

Practice A03

$$\displaystyle{\sum_{k=1}^{\infty}{\left[\frac{3^k}{(k+1)^k}\right]}}$$

solution

Practice A04

$$\displaystyle{\sum_{n=1}^{\infty}{\left[\frac{4n^2+1}{5n^2-8}\right]^n}}$$

solution

Practice A05

$$\displaystyle{\sum_{n=1}^{\infty}{\left[\frac{e^{3n}}{n^n}\right]}}$$

solution

 Level B - Intermediate

Practice B01

$$\displaystyle{\sum_{n=1}^{\infty}{\left[\frac{n^n}{3^{1+3n}}\right]}}$$

solution

Practice B02

$$\displaystyle{\sum_{n=1}^{\infty}{\left[\frac{n}{n+1}\right]^{n^2}}}$$

solution

Practice B03

$$\displaystyle{\sum_{n=1}^{\infty}{\left[\frac{(-1)^n(n^n)}{3^{n^3+1}}\right]}}$$

solution

Practice B04

$$\displaystyle{\sum_{n=1}^{\infty}{\left[\frac{(-1)^n}{[\tan^{-1}(n)]^n}\right]}}$$

$$\displaystyle{\sum_{n=1}^{\infty}{\left[\frac{n^{n-1}}{2^{3+n}}\right]}}$$