TeachingCalculus
- Sequences and series
- Finite sums and mathematical induction
- Sequences
- The Fibonacci sequence
- The Lucas sequence
- The integral test
- Comparison tests
- Alternating series
- Absolute convergence
- Power series
- Power series representations
- Remarkable decimal fractions
- A generating function for the Fibonacci numbers
- A generating function for the Lucas numbers
- Taylor series
- Catalan's constant
- The Riemann zeta function
- The harmonic numbers
- Pascal's triangle and the binomial theorem
- Binomial series
- Power series solutions of differential equations
Calculus – Sequences and series – Absolute convergence
Definition: A series \(\displaystyle\sum a_n\) is called absolutely convergent if the series of absolute values \(\displaystyle\sum|a_n|\) is convergent.
Theorem: If a series \(\displaystyle\sum a_n\) is absolutely convergent, then it is convergent.
Proof:
Observe that the inequality \(0\leq a_n+|a_n|\leq 2|a_n|\) is true because \(|a_n|\) is either \(a_n\) or \(-a_n\). If \(\displaystyle\sum a_n\)
is absolutely convergent, then \(\displaystyle\sum|a_n|\) is convergent, so \(\displaystyle\sum2|a_n|\) is convergent. Therefore, by the (direct)
comparison test \(\displaystyle\sum\left(a_n+|a_n|\right)\) is convergent. Then
is the difference of two convergent series and is therefore convergent.
Definition: A series \(\displaystyle\sum a_n\) is called conditionally convergent if it is convergent but not absolutely convergent.
Examples
- The alternating harmonic series \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\) is conditionally convergent,
- The series \(\displaystyle\sum_{n=1}^{\infty}\frac{\sin(n)}{n^2}\) is absolutely convergent.
The ratio test
Theorem:
- If \(\displaystyle\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L<1\), then the series \(\displaystyle\sum a_n\) is absolutely convergent,
- If \(\displaystyle\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L>1\) or \(\displaystyle\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\infty\), then the series \(\displaystyle\sum a_n\) is divergent,
- If \(\displaystyle\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1\), this test is inconclusive: the series \(\displaystyle\sum a_n\) might be convergent or might be divergent.
Proof:
If \(\displaystyle\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L<1\) we can choose a number \(r\) such that \(L < r < 1\). Then
\(\displaystyle\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L\) implies that there exists an integer \(N\) such that
\(\displaystyle\left|\frac{a_{n+1}}{a_n}\right| < r\) for all \(n\geq N\) or equivalently \(|a_{n+1}| < |a_n|r\) for all \(n\geq N\). Hence we have:
In general: \(|a_{N+k}| < |a_N|r^k\) for \(k=1,2,3,\ldots\). Now the series \(\displaystyle\sum_{k=1}^{\infty}|a_N|r^k\) is convergent because it is a geometric series with common ratio \(r\) and \(0 < r < r\). This implies that the series \(\displaystyle\sum_{n=N+1}^{\infty}|a_n|=\sum_{k=1}^{\infty}|a_{N+k}|\) is convergent. Hence the series \(\displaystyle\sum|a_n|\) is convergent.
If \(\displaystyle\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L>1\) or \(\displaystyle\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\infty\), then the ratio \(\displaystyle\left|\frac{a_{n+1}}{a_n}\right|\) will eventually be greater than \(1\), that is: there exists an integer \(N\) such that \(\displaystyle\left|\frac{a_{n+1}}{a_n}\right|>1\) for all \(n\geq N\). This implies that \(|a_{n+1}| > |a_n|\) for all \(n\geq N\) and therefore \(\lim\limits_{n\to\infty}a_n\neq0\). Hence the series \(\displaystyle\sum a_n\) is divergent.
In order to show that the test is inconclusive if \(\displaystyle\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1\), consider the divergent series \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}\) and the convergent series \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}\). Note that
\[\lim\limits_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}=\lim\limits_{n\to\infty}\frac{n}{n+1}=1\quad\text{and}\quad \lim\limits_{n\to\infty}\frac{\frac{1}{(n+1)^2}}{\frac{1}{n^2}}=\lim\limits_{n\to\infty}\frac{n^2}{(n+1)^2}=\lim\limits_{n\to\infty}\left(\frac{n}{n+1}\right)^2=1.\]Examples
1) Consider the alternating series \(\displaystyle\sum_{n=0}^{\infty}(-1)^n\frac{2n+1}{2^n}\). Let \(a_n=\displaystyle(-1)^n\frac{2n+1}{2^n}\), then we have:
\[\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim\limits_{n\to\infty}\left|(-1)^{n+1}\frac{2n+3}{2^{n+1}}\cdot(-1)^n\frac{2^n}{2n+1}\right| =\lim\limits_{n\to\infty}\frac{2n+3}{2n+1}\cdot\frac{1}{2}=\frac{1}{2}<1.\]This implies that the series is absolutely convergent.
2) Consider the series \(\displaystyle\sum_{n=0}^{\infty}\frac{n!}{5^n}\). Let \(a_n=\displaystyle\frac{n!}{5^n}\), then we have:
\[\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim\limits_{n\to\infty}\left|\frac{(n+1)!}{5^{n+1}}\cdot\frac{5^n}{n!}\right| =\lim\limits_{n\to\infty}\frac{1}{5}(n+1)=\infty.\]This implies that the series is divergent.
The root test
Theorem:
- If \(\displaystyle\lim\limits_{n\to\infty}\sqrt[n]{|a_n|}=L<1\), then the series \(\displaystyle\sum a_n\) is absolutely convergent,
- If \(\displaystyle\lim\limits_{n\to\infty}\sqrt[n]{|a_n|}=L>1\) or \(\displaystyle\lim\limits_{n\to\infty}\sqrt[n]{|a_n|}=\infty\), then the series \(\displaystyle\sum a_n\) is divergent,
- If \(\displaystyle\lim\limits_{n\to\infty}\sqrt[n]{|a_n|}=1\), this test is inconclusive: the series \(\displaystyle\sum a_n\) might be convergent or might be divergent.
The proof is similar to the proof of the ratio test and is based on the observation that if \(L < r < 1\) then \(\displaystyle\lim\limits_{n\to\infty}\sqrt[n]{|a_n|}=L\) implies that there exists an integer \(N\) such that \(\sqrt[n]{|a_n|} < r\) for all \(n\geq N\). This implies that
\[\sqrt[N+k]{|a_{N+k}|} < r\quad\Longrightarrow\quad |a_{N+k}| < r^{N+k},\quad k=0,1,2,\ldots.\]This implies that \(\displaystyle\sum_{k=0}^{\infty}|a_{N+k}|\) is convergent, since \(\displaystyle\sum_{k=0}^{\infty}r^{N=k}\) is a geometric series with common ratio \(r\) with \(0 < r < 1\) which is convergent.
Examples
1) Consider the series \(\displaystyle\sum_{n=0}^{\infty}\left(\frac{2n+1}{3n+4}\right)^n\). Let \(a_n=\displaystyle\left(\frac{2n+1}{3n+4}\right)^n\), then we have:
\[\lim\limits_{n\to\infty}\sqrt[n]{|a_n|}=\lim\limits_{n\to\infty}\left|\frac{2n+1}{3n+4}\right|=\frac{2}{3}<1.\]This implies that the series is absolutely convergent.
2) Consider the series \(\displaystyle\sum_{n=1}^{\infty}\left(\frac{2n+1}{n}\right)^n\). Let \(a_n=\displaystyle\frac{2n+1}{n}\), then we have:
\[\lim\limits_{n\to\infty}\sqrt[n]{|a_n|}=\lim\limits_{n\to\infty}\left|\frac{2n+1}{n}\right|=2.\]This implies that the series is divergent.
Last modified on March 15, 2021
