2.12 Tests for Convergence of Series
The Integral test
Let \(f(x)\) be a function such that \(f(n)\) is the general term of the series \(\displaystyle {\sum a_n}\) of positive terms. If \(f(x)>0\) and never increasing on the interval \(x> \alpha ,\) where \(\alpha \) is some positive integer, then the series \(\displaystyle {\sum a_n}\) converges or diverges according as \[\int ^{\infty }_{\alpha }f(x)\hspace {0.1cm}dx = \lim _{n \rightarrow +\infty }\int ^n_{\alpha }f(x)\hspace {0.1cm}dx\] exists or does not exist.
- 1.
- Examine the \(\displaystyle {\sum ^{\infty }_{n = 1}\frac {1}{n^2}}\) for convergent using the integral test.
- 2.
- Examine the \(\displaystyle {\sum ^{\infty }_{n = 1} \dfrac {1}{n}}\) for convergence using the integral test.
Solution.
Part 1
Here \(f(n) = a_n = \dfrac {1}{n^2}\), so we take \(f(x) = \dfrac {1}{x^2}\). On the interval \(x>1\),\(\hspace {0.2cm} f(x) >0\) and \(f(x)\) is decreasing as \(x\) increases. Thus we take \(\alpha = 1\). Thus, \begin {align*} \int _1^{+\infty }f(x)\hspace {0.1cm}dx & = \int _1^{+\infty }\frac {1}{x^2}dx = \lim _{n \rightarrow + \infty } \int _1^n \frac {1}{x^2}dx\\ & = \lim _{n \rightarrow + \infty }\Bigg [\dfrac {-1}{x}\Bigg ]^n_1\\\\ & = \lim _{n \rightarrow + \infty }\Bigg [\dfrac {-1}{n} + 1\Bigg ]\\\\ & = 1,\hspace {0.3cm}\text {which exist} \end {align*}
\(\therefore \hspace {0.4cm}\) by the integral test, \(\displaystyle {\sum a_n}\) converges.
Part 2
\(\displaystyle {f(n) = \frac {1}{n} \implies f(x) = \dfrac {1}{x}}\)
Similarly, on the integral \(x>1\), \(\hspace {0.1cm} f(x)>0\) and is decreasing as \(n\) increases. Taking \(\alpha = 1\), we have \begin {align*} \int _1^{\infty }f(x)\hspace {0.1cm}dx & = \int _1^{\infty }\frac {1}{x}\hspace {0.1cm}dx = \lim _{n\rightarrow \infty }\int _1^n \dfrac {1}{x}\hspace {0.1cm}dx\\ & = \lim _{n\rightarrow \infty }\Bigg [\ln |x|\Bigg ]^n_1\\ &= \lim _{n\rightarrow \infty }\Big [\ln n - \ln 1\Big ]\\ & = \lim _{n\rightarrow \infty }\Big [\ln n\Big ] = +\infty \\ \end {align*}
\(\implies \hspace {0.5cm}\displaystyle { \int _1^{\infty }f(x)\hspace {0.1cm}dx}\hspace {0.4cm}\) does not exist.
\(\therefore \hspace {0.5cm}\displaystyle {\sum ^{\infty }_{n=1}\dfrac {1}{n}\hspace {0.4cm}}\) diverges.
This series \(\displaystyle {\sum ^{\infty }_{n=1}\dfrac {1}{n}\hspace {0.4cm}}\) is called the harmonic series.
The Comparison test
A positive series \(\displaystyle {\sum a_n}\) is convergent if each terms (perhaps, after a finite number) is less than or equal to the corresponding term of a known convergent positive series \(\displaystyle {\sum a_n}\).
For comparison test the following series are useful as test series
- 1.
- The geometric series \[a + ar + ar^2 + \cdots \cdots \cdots \cdots + ar^n + \cdots \cdots \] for \(a\neq 0\), converges for \(0< r<1\) and diverges for \(r>1\).
- 2.
- The \(P\) series \(\hspace {0.3cm}\displaystyle {1 + \dfrac {1}{2^P} + \dfrac {1}{3^P} + \cdots \cdots \cdots + \dfrac {1}{n^P}+\cdots \cdots }\) which converges for \(P>1\), and diverges for \(P\leq 1\).
- 3.
- Each new series tested.
Example 2.12.2. Examine the \(\displaystyle {\sum ^{\infty }_{n = 1}\dfrac {1}{n^2 + 1}}\) for convergence using the comparison test.
Solution. The general term \(\dfrac {1}{n^2 + 1} < \dfrac {1}{n^2}\hspace {0.2cm} \forall n\). The series \(\displaystyle {\sum \dfrac {1}{n^2}}\) converges. By the comparison test the series \(\displaystyle {\sum ^{\infty }_{n = 1}\dfrac {1}{n^2 + 1}}\)
The Ratio Test
A positive series \(\displaystyle {\sum a_n}\) converges if \(\displaystyle { \lim \limits _{n\rightarrow + \infty } \frac {a_{n + 1}}{a_n}< 1}\), and diverges if \(\displaystyle { \lim \limits _{n\rightarrow + \infty } \frac {a_{n + 1}}{a_n}> 1}\) or if \(\displaystyle { \lim \limits _{n\rightarrow + \infty } \frac {a_{n + 1}}{a_n} = + \infty }\).
If \(\displaystyle { \lim \limits _{n\rightarrow + \infty } \frac {a_{n + 1}}{a_n}=1}\) then the test gives no information on convergence or divergence of the series.
Example 2.12.3. Examine the series \(\displaystyle {\sum _n^{\infty } \dfrac {n}{3^n}}\) for convergence using the ratio test.
Solution.
\(\displaystyle {a_n = \dfrac {n}{3^n} \hspace {0.3cm} \implies \hspace {0.3cm} a_{n+1} = \dfrac {n + 1}{3^{n+ 1}}}\)
\(\displaystyle { \frac {a_{n+1}}{a_n} = \frac {n+ 1}{3^{n + 1}} \times \frac {3^n}{n} = \frac {n+1}{3\times 3^n}\times \frac {3^n}{n}}\)
\(\displaystyle {\implies \hspace {0.5cm} \lim \limits _{n\rightarrow \infty } \frac {1}{3}\Bigg (1 + \frac {1}{n}\Bigg ) = \frac {1}{3}< 1}\)
By the ratio test, the series \(\displaystyle {\sum _n^{\infty } \dfrac {n}{3^n}}\) converges.
\(\star \hspace {0.5cm}\) A series having only negative terms can be treated the same way as a series of positive terms.
Definition 2.12.4. A series whose terms are alternatively positive and negative as \[\sum (-1)^{n-1} \hspace {0.1cm} a_n = a_1 - a_2 + a_3 + ----\hspace {0.2cm} (-1)^{n-1}a_n+ \hspace {0.2cm}\cdots \cdots \cdots \] is called an alternating series.
Convergence for alternating series
An alternating series \(\displaystyle {\sum (-1)^{n-1} \hspace {0.1cm} a_n}\) converges if
- \(i\).
- \(a_n > a_{n + 1}\)
- \(ii\).
- \(\lim \limits _{n \rightarrow \infty } a_n = 0\)
Example 2.12.5. Show that the series \(\displaystyle {\sum (-1)^{n-1}\hspace {0.1cm}\frac {1}{n}}\) converges.
Solution. \(\displaystyle {\sum (-1)^{n-1}\hspace {0.1cm}\frac {1}{n} = 1 - \frac {1}{2} + \frac {1}{3} - \frac {1}{4} + \cdots \cdots \cdots }\) is an alternating series.
- \(i\).
- \(1 > \dfrac {1}{2} > \dfrac {1}{3}>\hspace {0.2cm}\cdots \cdots \cdots \implies a_n > a_{n + 1}\)
- \(ii\).
- \(\lim \limits _{n \rightarrow \infty } a_n = \lim \limits _{n \rightarrow \infty } \dfrac {1}{n} = 0\)
By an alternating series test the series \(\displaystyle {\sum (-1)^{n-1}\hspace {0.1cm}\frac {1}{n}}\) converges.
Definition 2.12.6. A series \(\displaystyle {\sum a_n}\) with mixed (positive and negative) terms is called absolutely convergent if the series of absolute values \[\sum \left |a_n\right | = \left |a_1\right | + \left |a_2\right | + \left |a_3\right | + \cdots + \left |a_n\right | + \cdots \] converges.
If \(\displaystyle {\sum a_n}\) converges while \(\displaystyle {\sum \left |a_n\right |}\) diverges, then \(\displaystyle {\sum a_n}\) is said to be conditionally convergent.
Ratio Test for Absolutely Convergent series
A series \(\displaystyle {\sum a_n}\) with mixed terms is absolutely convergent if \[\lim \limits _{n\rightarrow \infty }\left |\dfrac {a_{n+1}}{a_n}\right | < 1 ,\] and is divergent if \[\lim \limits _{n\rightarrow \infty }\left |\dfrac {a_{n+1}}{a_n}\right | > 1 .\]
If \(\displaystyle {\lim \limits _{n\rightarrow \infty }\left |\dfrac {a_{n+1}}{a_n}\right | = 1}\hspace {0.2cm}\), test fails.
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