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Calculus An intuitive approach

Wai Yan Pong

Section 5.4 Convergence Tests for Series

Given the importance of series, there are a number of tests for it convergence. We only discuss a few basic ones in this section. Check out this article for more information.
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As we have seen, it is often easier to determine the convergence of a sequence than finding (meaning recognizing a nice expression) its limit. Likewise, it is often easier to determine the convergence of a series than finding its sum.
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The first series of results is based on the simple observation: for a series \(\sum a_n\) of non-negative real numbers, the sequences of partial sums clearly increasing. Hence it follows form the Monotonic Convergence Theorem 5.16 that
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By interpreting the sum of a series with non-negative terms as area, we can compare it with an improper integral.
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Proof.

Since \(f\) is decreasing and is positive on \([1,\infty)\text{,}\) for \(k \ge 1\text{,}\)
\begin{equation} \int_{k}^{k+1} f(x) dx \le f(k) \le \int_{k-1}^k f(x) dx\tag{5.1} \end{equation}
Summing \(k\) from \(2\) through \(n\) yields
\begin{equation} \int_{2}^{n+1} f(x) dx \le \sum_{n=2}^n f(k) \le \int_{1}^{n} f(x) dx. \tag{5.2} \end{equation}
Adding \(f(1)\) to each term above, we get
\begin{gather*} f(1) + \int_{2}^{n+1} f(x) dx \le \sum_{n=1}^n f(k) \le f(1) + \int_{1}^{n} f(x) dx. \\ \int_1^{n+1} f(x) dx \le s_n \le f(1) + \int_1^{n} f(x) dx \end{gather*}
where \(s_n\) is the \(n\)th partial sum of the series \(\sum f(k)\text{.}\) Thus, if the improper integral \(\int_1^\infty f(x) dx\) is convergent, then since \(f\) is positive on \([1,\infty)\text{,}\) \(\int_1^n f(x) dx\) is bounded above by this value for each \(n\text{.}\) And by the second inequality above \(s_n \le f(1) + \int_1^{\infty} f(x) dx\) for each \(n\text{.}\) Therefore, the series \(\sum_{k=1}^\infty f(k)\) is convergent (Proposition 5.36) moreover, its sum cannot excceed \(f(1)+\int_1^{\infty} f(x)dx \text{.}\)
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On the other hand if the improper integral is divergent, then \(\int_1^{n+1} f(x) dx\) diverges to \(+\infty\text{,}\) and so must the series \(\sum_{k=1}^\infty f(k)\) because of the first inequality above.
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As a bonus of the proof of the integral test, we get an upper bound of the error when the sum a series is approximated by a partial sum. Since
\begin{equation*} s_n - s_m = \sum_{k=m+1}^n f(k) \le \int_m^n f(x) dx, \end{equation*}
so when the improper integral \(\int_1^\infty f(x) dx\) is convergent we have, as \(n \to \infty\text{,}\)
\begin{equation*} R_m: s - s_m = \sum_{n=m+1}^\infty f(n) \le \int_m^\infty f(x) dx. \end{equation*}
The sum \(R_m\) is called the \(m\)th tail (or \(m\)th remainder) of the series \(\sum f(n)\text{.}\) It is the error when estimating a convergent series by its \(m\)th partial sum.
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Example 5.38.

It follows form the integral test and Proposition 4.3 that the Harmonic series, \(\sum_{n=1}^{\infty} \frac{1}{n}\text{,}\) is divergent and the series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) is convergent. It sums to \(\frac{\pi^2}{6}\) and this realization has an interesting history. See this page for the story and see this video for an interesting explanation.
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Taking absolute value of the terms of a series results in a series of non-negative numbers. This leads to the concept of absolute convergence. A series \(\sum a_n\) is absolutely convergent (or converges absolutely) if the series \(\sum |a_n|\) is convergent.
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For series of non-negative numbers, there is no difference between convergence and absolute convergence. In general, however, a convergence series needs not be absolutely convergent. We said that a series is conditionally convergent if it is convergent but not absolutely convergent.
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Example 5.40.

The series \(\sum \frac{(-1)^n}{n^2}\) is absolutely convergent and hence is convergent.
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Continuing the theme of comparison, we have the follow two results on convergence for series with positive terms.
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Proof.

Let \((s_k)\) and \((u_k)\) be the sequence of partial sums of \(\sum x_n\) and \(\sum y_n\) respectively. The conditions \(0 \le a_n \le b_n\) for all \(n\) imply \(0 \le s_k \le u_k\) for all \(k\text{.}\) The proposition now follows from Proposition 5.36
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Example 5.42.

The series \(\sum \frac{1}{n^2 + n+1}\) converges. This follows from the comparison test since \(0 \le \frac{1}{n^2+n+1} \le \frac{1}{n^2}\) and \(\sum \frac{1}{n^2}\) is convergent.
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The following test makes the comparison easier to apply in many cases.
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Before giving the proof, let us argue why the limit comparison test is intuitively clear. If the limit \(L\) of \((a_n/b_n)\) is non-zero, that means \(a_n\) is more or less \(L b_n\) and \(b_n\) is more or less \(a_n/L\text{.}\) Thus, the series \(\sum a_n\) and \(\sum b_n\) have the same convergence. If \(L=0\text{,}\) that means eventually, \(b_n\) is must bigger than \(a_n\) and so \(\sum a_n\) converges if \(\sum b_n\) does.
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Proof.

Since \(a_n/b_n \to L \gt 0\text{,}\) for all \(n\) sufficiently large,
\begin{equation*} \frac{L}{2} \lt \frac{a_n}{b_n} \lt \frac{3L}{2}. \end{equation*}
Since \(b_n \gt 0\text{,}\) that means for all \(n\) sufficiently large
\begin{equation*} \frac{L}{2}b_n \lt a_n \lt \frac{3L}{2}b_n. \end{equation*}
So the \(L \gt 0\) case follows from Proposition 5.41 and the fact that the convergence of a series of positive numbers is unaffected by multiplying the series by a positive constant. If \(L=0\text{,}\) then \(0 \lt a_n/b_n \lt 1\) and so \(0 \lt a_n \lt b_n\) for all \(n\) sufficiently large. Thus, the convergence of \(\sum a_n\) follows from the convergence of \(\sum b_n\) by comparison test as well.
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Example 5.44.

The series \(\sum \frac{1}{n^2 -n+1}\) is convergent. This follows from the limit comparison test since it terms are positive,
\begin{equation*} \frac{\frac{1}{n^2}}{\frac{1}{n^2-n+1}} = \frac{n^2-n+1}{n^2} = 1 - \frac{1}{n}+\frac{1}{n^2} \to 1 \end{equation*}
and \(\sum \frac{1}{n^2}\) is convergent.
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Proof.

Let \(r = (1+L)/2\text{.}\) If \(L \lt 1\text{,}\) then \(r \gt 1\text{.}\) Also, eventually,
\begin{equation*} 0 \lt \frac{|a_{n+1}|}{|a_n|} \lt r:=\frac{1+L}{2} \lt 1. \end{equation*}
That is \(|a_{n+1}| \lt |a_n|r\text{.}\) Since convergence is the only concern here, we can assume the inequality holds for all \(n \ge 1\text{.}\) Consequently,
\begin{equation*} |a_2| \lt |a_1|r, |a_3| \lt |a_2|r \lt |a_1|r^2, \ldots \end{equation*}
Since \(0 \lt r \lt 1\text{,}\) the geometric series \(|a_1|\sum r^n\) converges and so must \(\sum |a_n|\) by the Comparison Test.
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On the other hand, if \(L \gt 1\text{,}\) then so is \(r\) and
\begin{equation*} |a_{n}| > |a_1|r^{n-1}. \end{equation*}
Therefore, \((a_n)\) is not a null sequence and so \(\sum a_n\) diverges by the Divergence Test.
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Example 5.46.

The series \(\sum a_n = \sum \frac{-7^n}{n!}\) is absolutely convergent by Ratio test because
\begin{equation*} \frac{|a_{n+1}|}{|a_n|} = \frac{\frac{7^{n+1}}{(n+1)!}}{\frac{7^n}{n!}} = \frac{7^{n+1}}{(n+1)!}\frac{n!}{7^n}= \frac{7}{n+1} \to 0. \end{equation*}
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Proof.

Let \(r = (1+L)/2\text{.}\) If \(L \lt 1\text{,}\) then so is \(r\text{.}\) And a similar argument as in the proof of the Ratio test, we can assume for all \(n \ge 1\text{,}\)
\begin{equation*} |a_n|^{1/n} \lt r \end{equation*}
That is \(|a_n| \lt r^n\text{.}\) So we know that \(\sum |a_n|\) converges by comparing it with the geometric series \(\sum r^n\text{.}\)
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On the other hand, if \(L \gt 1\text{,}\) then so is \(r\text{.}\) Again by a similar argument as in the proof of the Ratio Test, we can assume
\begin{equation*} |a_n|^{1/n} \lt r \end{equation*}
and hence \(|a_n| > r^n\) for all \(n \ge 1\text{.}\) That means \((a_n)\) is not null and hence \(\sum a_n\) diverges.
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Example 5.48.

The series \(\sum \frac{2^n}{n^4 5^{n+1}}\) converges absolutely. To see this, note that the series is \(\frac{1}{5}\sum a_n\) where \(a_n = \frac{2^n}{n^4 5^n}\text{.}\) So the original series and the series \(\sum a_n\) have the same convergence. Note that
\begin{equation*} \left|\frac{2^n}{n^4 5^n}\right|^{1/n} = \frac{2}{n^{4/n}5} \to \frac{2}{5} \lt 1 \end{equation*}
because \(n^{1/n} \to 1\) Example 5.23. Thus, \(\sum a_n\) and hence the original series converges absolutely according to the root test.
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Remark 5.49.

The Ratio and the Root test are both inconclusive when the corresponding limit \(L\) is \(1\text{.}\) The series \(\sum 1/n\) and \(\sum 1/n^2\) both have \(L=1\) for the \(L\) in both the root and the ratio test, yet \(\sum 1/n^2\) converges absolutely but \(\sum 1/n\) diverges.
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There are finer versions the ratio and the root test involving limit superior.
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The root test is stronger than the ratio test in the sense that whenever the ratio test can decide the convergence of a series so does the root test but not the other way around.
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A series is alternating if the sign of its terms change from one to the next. That means it is a series of form \(\sum (-1)^{n+1} a_n\) or \(\sum (-1)^n a_n\) where \((a_n)\) is a sequence of positive numbers. Clearly, these two forms have the same convergence, so when discussing the convergence of an alternating series, we can assume its first term is positive.
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Proof.

Let \((s_k)\) be the sequence of partial sums of \(\sum (-1)^{n+1} a_n\text{.}\) Since \((a_n)\) is a strictly decreasing sequence of positive numbers, it follows that
\begin{equation*} s_2 \lt s_4 \lt s_6 \lt \cdots \lt s_5 \lt s_3 \lt s_1 \end{equation*}
Thus, \((s_{2m})\) is a strictly increasing sequence bounded above by \(s_1\) and \((s_{2m-1})\) is a strictly decreasing sequence bounded below by \(s_2\text{.}\) So, both of them converge, say \(s_{2m} \to L_0\) and \(s_{2m-1} \to L_1\text{.}\) As the limit of a strictly decreasing sequence \(L_1 \lt s_{2m-1}\) for all \(m\text{.}\) Likewise, \(L_0 \gt s_{2m}\) for all \(m\text{.}\) Moreover,
\begin{equation*} |L_1-L_0| \lt |s_{2m-1}-s_{2m}| = a_{2m}. \end{equation*}
By the assumption \(a_n \to 0\text{,}\) \(a_{2m} \to 0\) as well. Therefore, \(L_0 = L_1\text{.}\) Denote, this common value by \(L\) and we have \(L\) is always between any two consecutive partial sums and so for any \(k\text{,}\) \(|L-s_k| \lt |s_k-s_{k+1}| = a_{k+1}\text{.}\)
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Example 5.51.

The alternating series
\begin{equation*} \sum \frac{(-1)^{n+1}}{n} = 1-\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... \end{equation*}
satisfies the criteria in the alternating series test (with \(a_n = 1/n\)) and so is convergent. Since \(\sum |(-1)^{n+1}/n| = \sum 1/n\) is the Harmonic series, so \(\sum (-1)^{n+1}/n\) is a series that converges conditionally, i.e. it converges but not absolutely converges.
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Suppose \((a_n)\) is a sequence satisfying the criteria in the alternating series test. Then sum \(L\) of the series and the partial sums \(s_k\) satisfying \(|L-s_k| \le a_{k+1}\) for all \(k\text{,}\) we can use this information to compute, \(k\text{,}\) the number needed so that the \(k\)th partial sum \(s_k\) is within a given distant from \(L\text{.}\) Let see an example.
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Example 5.52.

We want to find a \(k\) so that \(s_k\) the \(k\)th partial sum of the alternating series \(\sum_{n=1}^{\infty} (-1)^{n-1} \frac{4}{n^4}\) is within given distance \(E \gt 0\) from the sum \(L\) of the series. Since the sequence \(a_n = 4/n^4\) satisfying the criteria in the alternating series test, the series converges. Moreover, we know that \(|L-s_k|\le a_{k+1}\) for each \(k\text{.}\) So all we need is to find a \(k\) such that \(a_{k+1} =\frac{4}{(k+1)^4} \le E\text{.}\) Solving for \(k\text{,}\) we find that we can take \(k\) to be the least integer so that
\begin{equation*} k \ge \left( \frac{4}{E} \right)^{1/4} -1. \end{equation*}
If a specific \(E\) is given, say \(E = 0.0006\) then we need
\begin{equation*} k \ge \left( \frac{4}{0.0006} \right)^{1/4} -1 \approx 8.03. \end{equation*}
So \(k=9\) will do. That is \(s_9 \approx 3.788370\) the \(9\)th partial sums of the series approximates the actual sum to within \(0.0006\text{.}\)
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