The **ratio test** and the **root test** are two more ways of checking for convergence of infinite series. The **ratio test** asks whether, in the limit that the number of terms goes to infinity $(n \rightarrow \infty)$, the ratio of the (**n+1**)^{th} term to the **n ^{th}** term is less than one.

The **root test** checks whether the limit, as $n \rightarrow \infty$, of the **n ^{th}** root of the

These tests (particularly the ratio test) are valuable in the next phase of learning about infinite series, when we determine what range of a variable x allows the series to converge – its *radius of convergence* or its *interval of convergence*.

First we have to take a little digression to refine our understanding of convergence of a series.

We have dealt with alternating series, and the criteria for convergence of those are somewhat easier than for non-alternating (positive) series. Here we'll take a deeper look at the meaning of convergence. First consider the series

$$\sum_{n=1}^{\infty} \, \frac{1}{n^2} \phantom{000} \color{#E90F89}{\text{converges absolutely}}$$

We've shown that this is a convergent p-series, with (p = 2) > 1. We can further claim that this series **converges absolutely** because

$$\text{If } \sum_{n = 1}^{\infty} |a_n| \; \text{ converges,} \\ \\ \text{ so does } \sum_{n = 1}^{\infty} a_n$$

This is called the absolute convergence test. It's worth thinking about a counterexample to understand absolute convergence. Consider the series

$$\sum_{n=1}^{\infty} \, \frac{(-1)^n}{n} \phantom{000} \color{#E90F89}{\text{converges conditionally}}$$

Under the absolute convergence test, this series fails, because taking the absolute value of the term just gives us the divergent harmonic series with terms 1/n. Yet we know that this *alternating* series does converge. We say that it **converges conditionally**, the condition being that the terms alternate sign.

The test for conditional convergence is

$$ \begin{align} &\sum_{n - 1}^{\infty} a_n \; \text{ converges conditionally} \\ \text{if } &\sum_{n = 1}^{\infty} a_n \text{ converges and } \sum_{n = 1}^{\infty} |a_n | \text{ diverges.} \end{align}$$

That is certainly the case for the alternating harmonic series.

The last case is a divergent series like the simple harmonic series:

$$\sum_{n=1}^{\infty} \, \frac{1}{n} \phantom{000} \color{#E90F89}{\text{diverges}}$$

A series like this diverges by one of our established convergence tests (in this case the integral test works well), or its terms simply do not decrease, so it fails the divergence test.

Test for **absolute convergence**

$$\text{If } \sum_{n=1}^{\infty} \, |a_n| \phantom{00} \text{converges, so does } \phantom{00} \sum_{n=1}^{\infty} \, a_n$$

Test for **conditional convergence**

$$ \sum_{n=1}^{\infty} \, a_n \phantom{00} \text{converges and} \phantom{00} \sum_{n=1}^{\infty} \, |a_n| \phantom{00} \text{diverges.}$$

**Divergence** – the series diverges by any one of the tests for convergence, or the divergence test.

The ratio test is a pretty straightforward test for convergence, and can distinguish between absolute and conditional convergence for series with negative terms.

We'll state it here, then do some examples. Proofs of the tests and theorems on this page will be saved for last.

$$\text{Let} \; \; L = \lim_{n\to\infty} \left| \frac{a_{n + 1}}{a_n} \right|$$

Then there are three possibilities for the limit, **L**:

If **L < 1****converges absolutely**

If **L > 1****diverges**

If **L = 1****inconclusive**

The following examples (and many ratio-test problems) make heavy use of factorials. If you're rusty, you might want to review their properties before moving on.

$$\text{Does } \; \sum_{n = 1}^{\infty} \frac{2^n}{(n + 1)!} \; \text{ converge?}$$

** Solution**: We'll set up the ratio test for this series and see how it goes. In setting up these ratios remember that

$$L = \lim_{n\to\infty} \left| \frac{a_{n + 1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{\frac{2^{n + 1}}{(n + 2)!}}{\frac{2^n}{(n + 1)!}} \right|$$

Now generally we'll get a single fraction by recognizing that division is multiplication by the reciprocal:

$$= \lim_{n\to \infty} \left| \frac{2^{n + 1}}{(n + 2)!} \cdot \frac{(n + 1)!}{2^n} \right|$$

Now we can use the properties of factorials and the laws of exponents to do some simplification. Recall that $2^{n+1} = 2^n \cdot 2^1$ and $(n+2)! = (n+2)(n+1)!$, so we have:

$$= \lim_{n\to \infty} \left| \frac{2^n 2^1 (n + 1)!}{(n + 2)(n + 1)! 2^n} \right|$$

which reduces to

$$= \lim_{n\to \infty} \left| \frac{2}{n + 2} \right| = 0 \lt 1$$

The limit is less than one, so the series converges. Hopefully, you had a pretty good suspicion that this series was convergent from the beginning, because the factorial function grows much more rapidly than an exponential function for $n \gt N$, some number at or beyond the point where the two functions cross.

*which* number the series converges to, just that it does. It doesn't give us the *sum* of the series. We would need to find that in another way.

$$\text{Does } \; \sum_{n = 1}^{\infty} \frac{e^n}{n} \; \text{ converge?}$$

** Solution**: First set up the ratio test:

$$L = \lim_{n\to \infty} \frac{e^{n + 1}}{n + 1} \cdot \frac{n}{e^n}$$

Now, recognizing that

$$\frac{e^{n + 1}}{e^n} = e^{(n + 1) - n} = e^1 = e$$

and using L'Hopital's rule to evaluate the limit, we get:

$$\lim_{n\to \infty} \frac{e\cdot n}{n + 1} = e \gt 1$$

The series diverges. Notice that this series wouldn't have passed the divergence test, either.

$$\text{Does } \; \sum_{n = 1}^{\infty} \frac{\sqrt{n}}{n^2 + 1} \; \text{ converge?}$$

** Solution**: We begin by setting up the ratio-test expression. I'm going to forgo the absolute value signs this time (just assume they're there), and I'll just write the denominator as a reciprocal product right away:

$$\lim_{n\to \infty} \frac{\sqrt{n + 1}}{(n + 1)^2 + 1} \cdot \frac{n^2 + 1}{\sqrt{n}}$$

Now expand the $(n+1)^2$ and rearrange to keep the roots together:

$$\lim_{n\to \infty} \frac{\sqrt{n + 1}}{\sqrt{n}} \cdot \frac{n^2 + 1}{n^2 + 2n + 3}$$

The properties of limits allow us to separate this expression into a product of limits:

$$\lim_{n\to \infty} \left( \frac{n + 1}{n} \right)^{1/2} \cdot \lim_{n\to \infty} \frac{n^2 + 1}{n^2 + 2n + 3}$$

The properties of limits also allow us to put the power of 1/2 on the left outside of the limit ...

$$= \left( \lim_{n\to\infty} \frac{n + 1}{n} \right)^{1/2} \cdot \lim_{n\to\infty} \frac{2n}{2n + 2}$$

... to obtain a limit of **1**, which means that the ratio test is

This series can be shown to be convergent by direct comparison to the p-series with **p = 3/2**.

Determine whether these series converge using the ratio test.

1. |
$$\sum_{n = 1}^{\infty} \frac{3^n}{n!}$$ |

2. |
$$\sum_{n = 1}^{\infty} \frac{a^n (n!)^2}{(2n)!} \; for \; a \gt 0$$ |

3. |
$$\sum_{n = 1}^{\infty} \frac{n^n}{n!}$$ |

4. |
$$\sum_{n = 1}^{\infty} \frac{(2n - 1)!}{(3n)!}$$ |

5. |
$$\sum_{n = 1}^{\infty} \frac{n! (n + 1)!}{(3n)!}$$ |

6. |
$$\sum_{n = 1}^{\infty} (-1)^n \frac{n!}{\pi^n}$$ |

In the ratio test, we consider the limit

$$L = \lim_{n\to\infty} \left| \frac{a_{n + 1}}{a_n} \right|,$$

In this proof, we'll look at two cases: L < 1 (convergence) and L > 1 (divergence).

Our strategy will be to compare our series, $\sum a_n$ with a convergent geometric series using the comparison test. If we set our ratioto be less than some value, $r$, we have:

$$\left| \frac{a_{n + 1}}{a_n} \right| \lt r$$

That is, for some value of $n \ge 0,$ let's call it $N,$ the ratio will be less than some number $r.$ This comes from the formal definition of limts.

Now that inequality can be rearranged by multiplying both sides by $|a_n|,$ like this:

$$|a_{n + 1}| \lt |a_n|\, r$$

Now if we let n be N (the value at which our limit is less than r), N+1, N + 2, and so on, we have

$$ \begin{align} |a_{n + 1}| &\lt |a_n|r \\[2pt] |a_{n + 2}| &\lt |a_{n + 1}|r \\[2pt] |a_{n + 3}| &\lt |a_{n + 2}|r \dots \end{align}$$

We have

$$|a_{N + k}| \lt |a_N| r^k \; \; \; for \; k \ge 1$$

Now the terms on the right of the inequality are those of a convergent geometric series. The comparison test says that if the terms of our series are smaller, term-by-term, than those of a convergent series, than it is also convergent.

This time, we're interested in the inequality

$$\left| \frac{a_{n + 1}}{a_n} \right| \gt 1$$

If this is true, then for some value of n, we'll call it N, the ratio will be greater than 1. So for $n \ge N, \; |a_{n + 1}| \gt |a_n|,$ which is the criterion for divergence using the divergence test. If the terms of a series grow, it cannot converge.

The root test is used in situations where a series term or part of it is raised to the power of the index variable. If the root test isn't fairly easy to use, you probably shouldn't use it. But when it works, it often cuts to convergence or divergence quickly.

The root test isn't a good choice if a series contains factorial terms. You'll notice some similarities between how we interpret the root test and the ratio test.

$$\text{Let} \; \; L = \lim_{n\to\infty} \left| a_n \right|^{\frac{1}{n}}$$

Then there are three possibilities for the limit, **L**:

Then there are three possibilities for the limit, **L**:

If **L < 1****converges absolutely**

If **L > 1****diverges**

If **L = 1****inconclusive**

$$\text{Does } \; \sum_{n = 1}^{\infty} \left( \frac{2n + 1}{5n - 3} \right)^n \; \text{ converge?}$$

** Solution**: This series is a perfect candidate for the root test because taking the n

$$\lim_{n\to\infty} \left| \left(\frac{2n + 1}{5n - 3} \right) \right|^{\frac{1}{n}} = \lim_{n\to\infty} \left( \frac{2n + 1}{5n - 3} \right)$$

Now we can use L'Hopital's rule to find the limit:

$$= \lim_{n\to\infty} \left( \frac{2n + 1}{5n - 3} \right) = \frac{2}{5} \lt 1$$

... so the series converges.

Series raised to a power of the index variable (the variable that counts in the summation) are good candidates for the root test.

$$\text{Does } \; \sum_{n = 1}^{\infty} \frac{(-3)^n}{n^n} \; \text{ converge?}$$

** Solution**: Here is another great candidate for the root test because of the powers of

$$\lim_{n\to\infty} \left| \left( \frac{(-3)}{n} \right)^n \right|^{\frac{1}{n}} = \lim_{n\to\infty} \frac{-3}{n} = 0$$

Because 0 < 1, the series converges. Notice that the series with terms -3/n would not converge (-3 times a harmonic series, but that's not what's going on here).

$$\text{Does } \; \sum_{n = 1}^{\infty} \frac{2^n n^3}{3^n} \; \text{ converge?}$$

** Solution**: First set up the limit. The

$$\lim_{n\to\infty} \left| \frac{2^n n^3}{3^n} \right|^{\frac{1}{n}} = \lim_{n\to\infty} \frac{2n^{\frac{3}{n}}}{3} = 0$$

Now the **n ^{3/n}** can be rewritten as

$$\lim_{n\to\infty} \frac{2}{3} \left( n^{\frac{1}{n}} \right)^3 = \frac{2}{3}$$

... so the series converges.

Determine whether these series converge using the root test.

1. |
$$\sum_{n = 1}^{\infty} \frac{2^{2n} 3^n}{10^n}$$ |

2. |
$$\sum_{n = 1}^{\infty} \frac{1}{n^2 2^n}$$ |

3. |
$$\sum_{n = 1}^{\infty} \frac{2^n}{6^n \sqrt{n}}$$ |

4. |
$$\sum_{n = 1}^{\infty} \left( \frac{1}{5} + \frac{1}{n} \right)^n$$ |

In the ratio test, we consider the limit

$$L = \lim_{n\to\infty} \left| a_n \right|^{\frac{1}{n}},$$

In this proof, we'll look at two cases: L < 1 (convergence) and L > 1 (divergence).

Choose some number $r$ between $L$ and 1, $L \lt r \lt 1.$ Then $\lim_{n\to\infty} \sqrt[n]{|a_k|} = L \lt r,$ so for large-enough $n,$ let's call it $n \ge N,$ it must be true that

$$\sqrt[n]{|a_n|} \le r,$$

which means that

$$|a_n| \le r^n$$

Now we've said that r < 1, so this geometric series,

$$\sum_{n = N}^{\infty} r^n = r^N + r^N \cdot r + r^N \cdot r^2 + \dots,$$

which converges to

$$\frac{r^N}{1 - r}.$$

So by the comparison test, $\sum |a_k|$ also converges because, term-by-term, its terms are less than those of a convergent geometric series, at least for $n \ge N.$

This time, we're interested in the inequality

$$\lim_{k\to\infty} \sqrt[n]{|a_n|} = L \gt 1$$

If this is true, then for some value of n that is large enough, say $n \ge N,$ it must be true that

$$\sqrt[n]{|a_n|} \ge 1,$$

which implies

$$|a_n| \ge 1^n = 1,$$

Thus this series diverges by the divergence test. If the terms of a series don't approach zero, the series cannot converge.

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