5.6 Ratio and root tests

In this section, we prove the last two series convergence tests: the ratio test and the root test. These tests are particularly nice because they do not require us to find a comparable series. The ratio test will be especially useful in the discussion of power series in the next chapter.

Throughout this chapter, we have seen that no single convergence test works for all series. Therefore, at the end of this section we discuss a strategy for choosing which convergence test to use for a given series.

Ratio test

Consider a series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}.$ From our earlier discussion and examples, we know that $\underset{n\to \infty }{\text{lim}}{a}_n=0$ is not a sufficient condition for the series to converge. Not only do we need $a_n\to 0,$ but we need $a_n\to 0$ quickly enough. For example, consider the series ${\displaystyle \sum _{n=1}^{\infty }1}\text{/}n$ and the series ${\displaystyle \sum _{n=1}^{\infty }1}\text{/}{n}^2.$ We know that $1\text{/}n\to 0$ and $1\text{/}{n}^2\to 0.$ However, only the series ${\displaystyle \sum _{n=1}^{\infty }1}\text{/}{n}^2$ converges. The series ${\displaystyle \sum _{n=1}^{\infty }1}\text{/}n$ diverges because the terms in the sequence $\left\{1\text{/}n\right\}$ do not approach zero fast enough as $n\to \infty .$ Here we introduce the ratio test    , which provides a way of measuring how fast the terms of a series approach zero.

Proof

Let ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ be a series with nonzero terms.

We begin with the proof of part i. In this case, $\rho =\underset{n\to \infty }{\text{lim}}\left|\frac{a_{n+1}}{a_n}\right|<1.$ Since $0\le \rho <1,$ there exists $R$ such that $0\le \rho <R<1.$ Let $\epsilon =R-\rho >0.$ By the definition of limit of a sequence, there exists some integer $N$ such that

Therefore,

and, thus,

Since $R<1,$ the geometric series

converges. Given the inequalities above, we can apply the comparison test and conclude that the series

converges. Therefore, since

where ${\displaystyle \sum _{n=1}^N|}{a}_n|$ is a finite sum and ${\displaystyle \sum _{n=N+1}^{\infty }|}{a}_n|$ converges, we conclude that ${\displaystyle \sum _{n=1}^{\infty }|}{a}_n|$ converges.

For part ii.

Since $\rho >1,$ there exists $R$ such that $\rho >R>1.$ Let $\epsilon =\rho -R>0.$ By the definition of the limit of a sequence, there exists an integer $N$ such that

Therefore,

and, thus,

Since $R>1,$ the geometric series

diverges. Applying the comparison test, we conclude that the series

diverges, and therefore the series ${\displaystyle \sum _{n=1}^{\infty }|}{a}_n|$ diverges.

For part iii. we show that the test does not provide any information if $\rho =1$ by considering the $p-\text{series}$ ${\displaystyle \sum }_{n=1}^{\infty }1\text{/}{n}^p.$ For any real number $p,$

However, we know that if $p\le 1,$ the $p-\text{series}$ ${\displaystyle \sum }_{n=1}^{\infty }1\text{/}{n}^p$ diverges, whereas ${\displaystyle \sum }_{n=1}^{\infty }1\text{/}{n}^p$ converges if $p>1.$

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The ratio test is particularly useful for series whose terms contain factorials or exponentials, where the ratio of terms simplifies the expression. The ratio test is convenient because it does not require us to find a comparative series. The drawback is that the test sometimes does not provide any information regarding convergence.