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Section 6.7 Ratio and Root Tests

Does the series \(\ds\sum_{n=0}^\infty {n^5\over 5^n}\) converge? It is possible, but a bit unpleasant, to approach this with the Integral Test or the Comparison Test, but there is an easier way. Consider what happens as we move from one term to the next in this series:

\begin{equation*} \cdots+{n^5\over5^n}+{(n+1)^5\over 5^{n+1}}+\cdots \end{equation*}

The denominator goes up by a factor of 5, \(\ds 5^{n+1}=5\cdot5^n\text{,}\) but the numerator goes up by much less: \(\ds (n+1)^5=n^5+5n^4+10n^3+10n^2+5n+1\text{,}\) which is much less than \(\ds 5n^5\) when \(n\) is large, because \(\ds 5n^4\) is much less than \(\ds n^5\text{.}\) So we might guess that in the long run it begins to look as if each term is \(1/5\) of the previous term. We have seen series that behave like this: The geometric series.

\begin{equation*} \sum_{n=0}^\infty {1\over 5^n} = {5\over4}\text{,} \end{equation*}

So we might try comparing the given series to some variation of this geometric series. This is possible, but a bit messy. We can in effect do the same thing, but bypass most of the unpleasant work.

The key is to notice that

\begin{equation*} \lim_{n\to\infty} {a_{n+1}\over a_n}= \lim_{n\to\infty} {(n+1)^5\over 5^{n+1}}{5^n\over n^5}= \lim_{n\to\infty} {(n+1)^5\over n^5}{1\over 5}=1\cdot {1\over5} ={1\over 5}\text{.} \end{equation*}

This is really just what we noticed above, done a bit more formally: in the long run, each term is one fifth of the previous term. Now pick some number between \(1/5\) and \(1\text{,}\) say \(1/2\text{.}\) Because

\begin{equation*} \lim_{n\to\infty} {a_{n+1}\over a_n}={1\over5}\text{,} \end{equation*}

then when \(n\) is big enough, say \(n\ge N\) for some \(N\text{,}\)

\begin{equation*} {a_{n+1}\over a_n}\lt {1\over2} \hbox{so} a_{n+1}\lt {a_n\over2}\text{.} \end{equation*}

So \(\ds a_{N+1}\lt a_N/2\text{,}\) \(\ds a_{N+2}\lt a_{N+1}/2\lt a_N/4\text{,}\) \(\ds a_{N+3}\lt a_{N+2}/2\lt a_N/8\text{,}\) and so on. The general form is \(\ds a_{N+k}\lt a_N/2^k\text{.}\) So if we look at the series

\begin{equation*} \sum_{k=0}^\infty a_{N+k}= a_N+a_{N+1}+a_{N+2}+a_{N+3}+\cdots+a_{N+k}+\cdots\text{,} \end{equation*}

its terms are less than or equal to the terms of the sequence

\begin{equation*} a_N+{a_N\over2}+{a_N\over4}+{a_N\over8}+\cdots+{a_N\over2^k}+\cdots= \sum_{k=0}^\infty {a_N\over 2^k} = 2a_N\text{.} \end{equation*}

So by the Comparison Test, \(\ds\sum_{k=0}^\infty a_{N+k}\) converges, and this means that \(\ds\sum_{n=0}^\infty a_{n}\) converges, since we've just added the fixed number \(\ds a_0+a_1+\cdots+a_{N-1}\text{.}\)

Under what circumstances could we do this? The crucial part was that the limit of \(\ds a_{n+1}/a_n\text{,}\) say \(L\text{,}\) was less than 1 so that we could pick a value \(r\) so that \(L\lt r\lt 1\text{.}\) The fact that \(L\lt r\) (in our example \(1/5\lt 1/2\)) means that we can compare the series \(\sum a_n\) to \(\sum r^n\text{,}\) and the fact that \(r\lt 1\) guarantees that \(\sum r^n\) converges. That's really all that is required to make the argument work.

The example above essentially proves the first part of this, if we simply replace \(1/5\) by \(L\) and \(1/2\) by \(r\text{.}\) Suppose that \(L>1\text{,}\) and pick \(r\) so that \(1\lt r\lt L\text{.}\) Then for \(n\ge N\text{,}\) for some \(N\text{,}\)

\begin{equation*} {|a_{n+1}|\over |a_n|} > r \hbox{and} |a_{n+1}| > r|a_n|\text{.} \end{equation*}

This implies that \(\ds |a_{N+k}|>r^k|a_N|\text{,}\) but since \(r>1\) this means that \(\ds\lim_{k\to\infty}|a_{N+k}|\not=0\text{,}\) which means also that \(\ds\lim_{n\to\infty}a_n\not=0\text{.}\) By the Divergence Test, the series diverges.

To see that we get no information when \(L=1\text{,}\) we need to exhibit two series with \(L=1\text{,}\) one that converges and one that diverges. The series \(\sum 1/n^2\) and \(\sum 1/n\) provide a simple example.

Note:

  1. The Ratio Test is particularly useful for series involving factorials and exponentials.

  2. Absolute Convergence:

    • In general, we require a series to just have non-zero terms.

    • Then we consider the absolute values of the terms: \(\lim\limits_{n\to\infty} \left\vert \dfrac{a_{n+1}}{a_n}\right\vert = L\)

    • This means we are testing for absolute convergence.

Example 6.60. Factorials and Ratio Test.

Analyze \(\ds\sum_{n=0}^\infty \frac{5^n}{n!}\text{.}\)

Solution
\begin{equation*} \lim_{n\to\infty} {5^{n+1}\over (n+1)!}{n!\over 5^n}= \lim_{n\to\infty} {5^{n+1}\over 5^n}{n!\over (n+1)!}= \lim_{n\to\infty} {5}{1\over (n+1)}=0\text{.} \end{equation*}

Since \(0\lt 1\text{,}\) the series converges.

A similar argument to the one used for the Ratio Test justifies a related test that is occasionally easier to apply, namely the so-called Root Test .

The proof of the Root Test is actually easier than that of the Ratio Test, and is left as an exercise.

Example 6.62. Exponentials and Root Test.

Analyze \(\ds\sum_{n=0}^\infty {5^n\over n^n}\text{.}\)

Solution

The Ratio Test turns out to be a bit difficult on this series (try it). Using the Root Test:

\begin{equation*} \lim_{n\to\infty} \left({5^n\over n^n}\right)^{1/n}= \lim_{n\to\infty} {(5^n)^{1/n}\over (n^n)^{1/n}}= \lim_{n\to\infty} {5\over n}=0\text{.} \end{equation*}

Since \(0\lt 1\text{,}\) the series converges.

Note:

  1. The Root Test is frequently useful for series involving exponentials.

  2. Absolute Convergence:

    • In general, we require a series to just have non-zero terms.

    • Then we consider the absolute values of the terms: \(\lim\limits_{n\to\infty} \left(\left\vert a_n \right\vert\right)^{1/n} = L\)

    • This means we are testing for absolute convergence.

Exercises for Section 6.7.

Compute the following limit for the given series.

  1. \(\ds\lim_{n\to\infty} |a_{n+1}/a_n|\) for \(\sum 1/n^2\)

    Answer
    1
    Solution
    We compute the limit:
    \begin{equation*} \lim_{n\to\infty} \frac{n^2}{(n+1)^2} \Heq \lim_{n\to\infty} \frac{2n}{2n} = 1. \end{equation*}

  2. \(\ds\lim_{n\to\infty} |a_{n+1}/a_n|\) for \(\sum 1/n\)

    Answer
    1
    Solution
    We compute the limit: \[\lim_{n\to\infty} \frac{n}{(n+1)} \Heq \lim_{n\to\infty} \frac{1}{1} = 1.\]

  3. \(\ds\lim_{n\to\infty} |a_n|^{1/n}\) for \(\sum 1/n^2\)

    Answer
    1
    Solution
    We compute the limit:
    \begin{equation*} \lim_{n\to\infty} \left(n^2\right)^{1/n} = \lim_{n\to\infty} n^{2/n}. \end{equation*}
    Now consider taking the log of this limit:
    \begin{equation*} \begin{split} \log\left( \lim_{n\to\infty} n^{2/n} \right) \amp= \lim_{n\to\infty}\log\left(n^{2/n}\right)\\ \amp= \lim_{n\to\infty} \frac{2}{n} \log(n) \\ \amp= 0\end{split} \end{equation*}
    Therefore,
    \begin{equation*} \lim_{n\to\infty} n^{2/n} = e^{0} = 1. \end{equation*}

  4. \(\ds\lim_{n\to\infty} |a_n|^{1/n}\) for \(\sum 1/n\)

    Answer
    1
    Solution
    We compute the limit:
    \begin{equation*} \lim_{n\to\infty} \left(n\right)^{1/n}. \end{equation*}
    Now consider taking the log of this limit:
    \begin{equation*} \begin{split} \log\left( \lim_{n\to\infty} n^{1/n} \right) \amp= \lim_{n\to\infty}\log\left(n^{1/n}\right)\\ \amp= \lim_{n\to\infty} \frac{1}{n} \log(n) \\ \amp= 0\end{split} \end{equation*}
    Therefore,
    \begin{equation*} \lim_{n\to\infty} n^{1/n} = e^{0} = 1. \end{equation*}

Determine whether the series converge.

  1. \(\ds\sum_{n=0}^\infty (-1)^{n}{3^n\over 5^n}\)

    Answer
    converges absolutely
    Solution
    We use the Root Test:
    \begin{equation*} \lim_{n\to\infty} \left\vert\frac{3^n}{5^n}\right\vert^{1/n} = \lim_{n\to\infty} \frac{3}{5} = \frac{3}{5} \lt 1. \end{equation*}
    Therefore, the series converges absolutely.

  2. \(\ds\sum_{n=1}^\infty {n!\over n^n}\)

    Answer
    converges
    Solution

    We use the Ratio Test:

    \begin{equation*} \begin{split} \lim_{n\to\infty} \left(\frac{(n+1)!}{(n+1)^{n+1}} \frac{n^n}{n!}\right) \amp = \lim_{n\to\infty} \left(\frac{n^n}{(n+1)^{n+1}} \frac{(n+1)!}{n!}\right)\\ \amp = \lim_{n\to\infty} \left(\frac{n^n}{(n+1)^{n+1}} (n+1)\right)\\ \amp = \lim_{n\to\infty} \left(\frac{n^n}{(n+1)^n}\right)\\ \amp = \lim_{n\to\infty} \frac{1}{(1+\frac{1}{n})^n} = \frac{1}{e}. \end{split} \end{equation*}

    Since \(\frac{1}{e} \lt 1\text{,}\) we conclude that the series

    \begin{equation*} \sum_{n=0}^{\infty} \frac{n!}{n^n} \end{equation*}

    converges absolutely.

  3. \(\ds\sum_{n=1}^\infty {n^5\over n^n}\)

    Answer
    converges
    Solution
    Since
    \begin{equation*} \lim_{n\to\infty} \left(\frac{n^5}{n^n}\right)^{1/n} = \lim_{n\to\infty} n^{(5-n)/n} = 0, \end{equation*}
    by the Root Test, the series converges.

  4. \(\ds\sum_{n=1}^\infty {(n!)^2\over n^n}\)

    Answer
    diverges
    Solution
    We use the Ratio Test:
    \begin{equation*} \begin{split} \lim_{n\to\infty} \left(\frac{(n+1)!^2}{(n+1)^{n+1}} \frac{n^n}{n!^2}\right) \amp= \lim_{n\to\infty} \left(\frac{n^n}{(n+1)^{n+1}} \frac{(n+1)!^2}{n!^2}\right)\\ \amp= \lim_{n\to\infty} \left(\frac{n^n}{(n+1)^{n+1}} (n+1)^2\right)\\ \amp= \lim_{n\to\infty} \left(\frac{n^n}{(n+1)^(n-1)}\right) \end{split} \end{equation*}
    Now take the logarithm of this limit:
    \begin{equation*} \log\left( \lim_{n\to\infty} \left(\frac{n^n}{(n+1)^(n-1)}\right)\right) = \lim_{n\to\infty} n\log n - (n-1)\log(n+1) = \infty. \end{equation*}
    Therefore, the series diverges.
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