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## Ratio and Root Tests

In this section we learn two tests for convergence of series: the Root Test and the Ratio Test.

Ratio Test: Consider a series $$\displaystyle\sum_{n=1}^{\infty} a_n$$. Let $L=\lim_{n\to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert.$

1. If $$L<1$$, then $$\displaystyle\sum_{n=1}^{\infty} a_n$$ is absolutely convergent.
2. If $$L>1$$, then $$\displaystyle\sum_{n=1}^{\infty} a_n$$ is divergent.
3. If $$L=1$$, then this test is inconclusive.

(a) Suppose $$L < 1$$. Let $$r$$ be a number such that $$L < r < 1$$. Since $$\displaystyle\lim_{n\to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert=L < r$$, there is a positive integer $$K$$ such that $\left\vert \frac{a_{n+1}}{a_n} \right\vert < r, \text{ i.e., } |a_{n+1}| < r|a_n| \text{ for all } n\geq K.$ Successively applying the last inequality, we get $|a_{K+n}| < r^n|a_K| \text{ for all } n\geq 1.$ Since $$\displaystyle \sum_{n=1}^{\infty} r^n|a_K|$$ is convergent as a geometric series, $$\displaystyle \sum_{n=1}^{\infty} |a_{K+n}|$$ is also convergent by the Comparison Test. Then so is the following as the sum of a finite number and a convergent series: $\sum_{n=1}^{\infty} |a_n| = \sum_{n=1}^{K} |a_n| +\sum_{n=1}^{\infty} |a_{K+n}|.$ Thus $$\displaystyle\sum_{n=1}^{\infty} a_n$$ is absolutely convergent.
(b) Suppose $$\displaystyle\lim_{n\to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert=L>1$$. Then there is a positive integer $$K$$ such that $\left\vert \frac{a_{n+1}}{a_n} \right\vert>1, \text{ i.e., } |a_{n+1}| > |a_n| \text{ for all } n\geq K.$ Then $$\displaystyle\lim_{n\to \infty} a_n\neq 0$$. Thus $$\displaystyle\sum_{n=1}^{\infty} a_n$$ is divergent by the Divergence Test.
(c) When $$L=1$$, $$\displaystyle\sum_{n=1}^{\infty} a_n$$ may be divergent (e.g., $$\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$$), absolutely convergent (e.g., $$\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$$), and conditionally convergent (e.g., $$\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$$).

Example.

1. Use the Ratio Test to determine if $$\displaystyle \sum_{n=1}^{\infty} \frac{n!}{n^n}$$ is absolutely convergent or divergent.

Solution. Here $$a_n=\displaystyle\frac{n!}{n^n}$$. Then $$a_{n+1}=\displaystyle\frac{(n+1)!}{(n+1)^{n+1}}$$ and \begin{align*} L &=\lim_{n\to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert \\ &=\lim_{n\to \infty} \left\vert \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} \right\vert \\ &=\lim_{n\to \infty} \frac{(n+1)!}{(n+1)^{n+1}} \frac{n^n}{n!} \\ &=\lim_{n\to \infty} \frac{(n+1)\cdot n!}{(n+1)\cdot (n+1)^n} \frac{n^n}{n!} \\ &=\lim_{n\to \infty} \left( \frac{n}{n+1} \right)^n\\ &= \frac{1}{\displaystyle\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^n} \\ &= \frac{1}{e}. \end{align*} Since $$L=\frac{1}{e}<1$$, $$\displaystyle \sum_{n=1}^{\infty} \frac{n!}{n^n}$$ is absolutely convergent by the Ratio Test and hence convergent.

2. Use the Ratio Test to determine if $$\displaystyle \sum_{n=1}^{\infty} \frac{2^n}{n^2}$$ is absolutely convergent or divergent.

Solution. Here $$a_n=\displaystyle\frac{2^n}{n^2}$$. Then $$a_{n+1}=\displaystyle\frac{2^{n+1}}{(n+1)^2}$$ and \begin{align*} L &=\lim_{n\to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert \\ &=\lim_{n\to \infty} \left\vert \frac{\frac{2^{n+1}}{(n+1)^2}}{\frac{2^n}{n^2}} \right\vert \\ &=\lim_{n\to \infty} \frac{2^{n+1}}{(n+1)^2} \frac{n^2}{2^n} \\ &=\lim_{n\to \infty} \frac{2\cdot 2^n}{2^n} \left(\frac{n}{n+1}\right)^2 \\ &=\lim_{n\to \infty} 2 \left(\frac{1}{1+\frac{1}{n}}\right)^2\\ &= 2 \left(\lim_{n\to \infty}\frac{1}{1+\frac{1}{n}}\right)^2 \\ &= 2. \end{align*} Since $$L=2>1$$, $$\displaystyle \sum_{n=1}^{\infty} \frac{2^n}{n^2}$$ is divergent by the Ratio Test.

3. Use the Ratio Test to determine if $$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}$$ is absolutely convergent or divergent.

Solution. Here $$a_n=\displaystyle \frac{1}{n^2}$$. Then $$a_{n+1}=\displaystyle \frac{1}{(n+1)^2}$$ and \begin{align*} L &=\lim_{n\to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert \\ &=\lim_{n\to \infty} \frac{1}{(n+1)^2} \frac{n^2}{1} \\ &=\lim_{n\to \infty} \left(\frac{1}{1+\frac{1}{n}}\right)^2\\ &= \left(\lim_{n\to \infty}\frac{1}{1+\frac{1}{n}}\right)^2 \\ &= 1. \end{align*} Since $$L= 1$$, the Ratio Test is inconclusive (although $$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}$$ is convergent by the $$p$$-series Test). Similarly we can show for $$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$$ that $$L=1$$ and the Ratio Test is inconclusive where the series is known to be divergent.

Root Test: Consider a series $$\displaystyle\sum_{n=1}^{\infty} a_n$$. Let $L=\lim_{n\to \infty}\displaystyle \left\vert a_n \right\vert^{\frac{1}{n}}.$

1. If $$L < 1$$, then $$\displaystyle\sum_{n=1}^{\infty} a_n$$ is absolutely convergent.
2. If $$L > 1$$, then $$\displaystyle\sum_{n=1}^{\infty} a_n$$ is divergent.
3. If $$L=1$$, then this test is inconclusive.

The proof of the Root Test is similar to that of the Ratio Test.

Example. Use the Root Test to determine if $$\displaystyle \sum_{n=1}^{\infty} \left( \frac{-2n}{n+1}\right)^{3n}$$ is absolutely convergent or divergent.

Solution. Here $$a_n=\displaystyle (-1)^{3n}\left(\frac{2n}{n+1} \right)^{3n}$$. Then $$|a_n|=\displaystyle \left(\frac{2n}{n+1} \right)^{3n}$$ and \begin{align*} L&=\lim_{n\to \infty} \left\vert a_n \right\vert^{\frac{1}{n}}\\ &=\lim_{n\to \infty} \left[ \left(\frac{2n}{n+1} \right)^{3n} \right]^{\frac{1}{n}} \\ &=\lim_{n\to \infty} \left(\frac{2n}{n+1} \right)^{3}\\ &= \left( \lim_{n\to \infty} \frac{2}{1+\frac{1}{n}} \right)^{3}\\ &= 2^3\\ &= 8. \end{align*} Since $$L=8 > 1$$, $$\displaystyle \sum_{n=1}^{\infty} \left( \frac{-2n}{n+1}\right)^{3n}$$ is divergent by the Root Test.

Note that this problem can also be done by the Ratio Test but the calculation will be a little bit more complicated:
Here $$a_n=\displaystyle (-2)^{3n}\left(\frac{n}{n+1} \right)^{3n}$$. Then $$a_{n+1}=\displaystyle(-2)^{3n+3}\left(\frac{n+1}{n+2} \right)^{3n+3}$$ and \begin{align*} L &=\lim_{n\to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert \\ &=\lim_{n\to \infty} \left\vert (-2)^{3n+3}\left(\frac{n+1}{n+2} \right)^{3n+3} \frac{1}{(-2)^{3n}}\left(\frac{n+1}{n} \right)^{3n} \right\vert \\ &=\lim_{n\to \infty} \left\vert -8\left(\frac{1+\frac{1}{n}}{1+\frac{2}{n}} \right)^{3n+3} \left(1+\frac{1}{n} \right)^{3n} \right\vert \\ &=\lim_{n\to \infty} 8\frac{\left(1+\frac{1}{n}\right)^{3}\left[\left(1+\frac{1}{n} \right)^{n}\right]^3}{\left(1+\frac{2}{n}\right)^{3}\left[\left(1+\frac{2}{n} \right)^{n}\right]^3} \left[\left(1+\frac{1}{n} \right)^{n}\right]^3 \\ &= 8 \frac{1^3\cdot e^3}{1^3\cdot (e^2)^3} e^3 \;\;\left( \text{since } \lim_{n\to \infty} \left(1+\frac{x}{n}\right)^n=e^x \right) \\ &= 8. \end{align*} Since $$L=8 > 1$$, $$\displaystyle \sum_{n=1}^{\infty} \left( \frac{-2n}{n+1}\right)^{3n}$$ is divergent by the Ratio Test.

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