Quotient Criterion
An important criterion to prove absolute convergence by means of ratios of the underlying sequence's terms.
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Study Quotient Criterion
Theorem 1 (Quotient criterion). Let \(n_0\in\mathbb{N}\) and let \(\sum_{k=1}^\infty a_k\) be a series in \(\mathbb{K}\) with the following properties:
\(a_k\neq0\) for all \(k\geq n_0\);
there exists some \(q\in(0,1)\) such that for all \(k\geq n_0\) holds \[\frac{|a_{k+1}|}{|a_{k}|}\leq q.\]
Then \(\sum_{k=1}^\infty a_k\) converges absolutely.
Proof: We inductively obtain that for all \(k\geq n_0\) holds \[|a_k|\leq q^{k-n_0}|a_{n_0}| \ .\] Therefore, the series \(\sum_{k=1}^\infty q^{k-n_0}|a_{n_0}|\) is a majorant of \(\sum_{k=1}^\infty a_k\). However, the majorant is convergent due to \[\sum_{k=1}^\infty q^{k-n_0}|a_{n_0}|=\frac{|a_{n_0}|}{(1-q)q^{n_0-1}}.\] The majorant criterion then implies convergence.\(\Box\)
Remark 2. Note that the quotient criterion is
different from claiming \(\frac{|a_{k+1}|}{|a_{k}|}<1\) (which is
for instance fulfilled by the divergent harmonic series). There has to
exist some \(q<1\) such that the
quotient is below \(q\).
The quotient criterion is only sufficient for convergence and indeed,
there are examples of absolutely convergent series that do not fulfill
the quotient criterion. For instance, consider the absolutely convergent
series \[\sum_{k=1}^\infty
\frac1{k^2}.\] Observing that \[\frac{|a_{k+1}|}{|a_{k}|}=\left|\frac{k}{k+1}\right|^2,\]
the fact that this expression converges to \(1\) implies that there does not exist some
\(q<1\) for which the quotient
criterion is fulfilled. However, this series is convergent as we have
proven in the Example on the geometric series.
We could also formulate “an alternative quotient criterion” that gives us a sufficient criterion for divergence. Namely, consider a real series \(\sum_{k=1}^\infty a_k\) with positive \(a_k\) and assume that \(\frac{a_{k+1}}{a_{k}}\geq 1\) for all \(k\geq n_0\) for some fixed \(n_0\in\mathbb{N}\). This gives us \(0< a_{k}\leq a_{k+1}\), i.e., the sequence \((a_n)_{n\in\mathbb{N}}\) is positive and monotonically increasing. Such a sequence cannot converge to zero and thus, \(\sum_{k=1}^\infty a_k\) is divergent.
Now we present a “limit form” of the quotient criterion:
Theorem 3. Quotient criterion (limit form) Let \(\sum_{k=1}^\infty a_k\) be a series in \(\mathbb{K}\) and assume that there exists some \(n_0\in \mathbb{N}\) such that \(a_k\neq0\) for all \(k\geq n_0\). If\[\limsup_{k \rightarrow \infty} \frac{|a_{k+1}|}{|a_{k}|}<1,\] then \(\sum_{k=1}^\infty a_k\) converges absolutely.
Proof: Set \(c:=\lim\sup\frac{|a_{k+1}|}{|a_{k}|}<1\). Since \(\lim\sup\) is defined to be the largest accumulation point of a sequence, we have for every \(\varepsilon>0\) that \(\frac{|a_{k+1}|}{|a_{k}|}\geq c+\varepsilon\) holds true for at most finitely many \(k\in\mathbb{N}\). Hence, for \(\varepsilon:=\frac{1-c}2,\) there exists some \(N\in\mathbb{N}\) such that for all \(k\geq N\) holds \[\frac{|a_{k+1}|}{|a_{k}|}<c+\varepsilon=c+\frac{1-c}2=\frac{1+c}2.\] Thus, the quotient criterion holds true for \(q:=\frac{1+c}2\) which satisfies \(q<1\) due to \(c<1\).
\(\Box\)
Remark 4. Since, in case of convergence of \(\frac{|a_{k+1}|}{|a_{k}|}\), the limit and limes superior coincide, the criterion \[\lim_{k\to\infty}\frac{|a_{k+1}|}{|a_{k}|}<1\] is also sufficient for absolute convergence of \(\sum_{k=1}^\infty a_k\). However, this criterion requires the convergence of the quotient sequence and is therefore weaker than the above one.
Example 5. For \(x\in\mathbb{K}\), consider the series \[\sum_{k=0}^\infty\frac{x^k}{k!}.\] Then an application of the limit form of the quotient criterion yields \[\,\limsup_{k \rightarrow \infty}\frac{\left|\frac{x^{k+1}}{(k+1)!}\right|}{\left|\frac{x^k}{k!}\right|}\] \[=\limsup_{k \rightarrow \infty}\frac{|x|^{k+1}}{|x|^k}\frac{k!}{(k+1)!}\] \[=\,\limsup_{k \rightarrow \infty}|x|\frac1{k+1} =\,\lim_{k\to\infty}|x|\frac1{k+1}=0<1.\] Hence, the series converges absolutely. The series \(\sum_{k=0}^\infty\frac{x^k}{k!}\) is called exponential series and we will indeed define \(\exp(x)\) by this expression.
Theorem 6. Let \(x\in\mathbb{K}\). Then \[\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n} = \sum_{k=0}^{\infty}\frac{x^{k}}{k!}.\]
Proof: First we show that for fixed \(k\in\mathbb{N}_0\) holds \[\begin{aligned}
\label{seqexp}
\lim_{n\rightarrow \infty} \frac{n!}{(n-k)!n^k} = 1.
\end{aligned}\] Note
that \(\frac{n!}{(n-k)!n^k}\) is well
defined for \(n\geq k\). For such an
\(n\in\mathbb{N}\) we have \[\begin{aligned}
1\geq \prod_{j=0}^{k-1} \frac{n-j}{n} = \frac{n!}{(n-k)!n^k}\geq
\frac{(n-k+1)^k}{n^k}=(1-\frac{k-1}{n})^k \ .
\end{aligned}\] Since the right-hand side converges to \(1\) for \(n\rightarrow \infty\), the claim of the
theorem follows.
Now let \(\varepsilon>0\) and let
\(K\in\mathbb{N}\) be big enough such
that \[\sum_{k=K}^{\infty}\frac{|x|^{k}}{k!}<\frac{\varepsilon}{3}.\]
By the convergence of the auxiliary sequence \(\frac{n!}{(n-k)!n^k}\), there is an \(N\geq K\) such that for all \(n\geq N\) also \[\sum_{k=0}^{K-1}\left|\frac{n!}{(n-k)!n^k}-1\right|\frac{|x|^{k}}{k!}
<\frac{\varepsilon}{3} \ .\]
For \(n\geq N\) we estimate \[\left|\left(1+\frac{x}{n}\right)^{n}-\sum_{k=0}^{\infty}\frac{x^{k}}{k!}\right| = \left|\sum_{k=0}^{n}\binom{n}{k}\frac{x^{k}}{n^{k}}-\sum_{k=0}^{\infty}\frac{x^{k}}{k!} \right|\] \[\leq \sum_{k=0}^{K-1}\left|\binom{n}{k}\frac{x^{k}}{n^{k}}-\frac{x^{k}}{k!}\right|+\sum_{k=K}^{n} \binom{n}{k}\frac{|x|^{k}}{n^{k}} +\underbrace{\sum_{k=K}^{\infty}\frac{|x|^{k}}{k!}}_{<\,\varepsilon/3}.\] \[= \underbrace{\sum_{k=0}^{K-1}\left|\frac{n!}{(n-k)!n^k}-1\right|\frac{|x|^{k}}{k!}}_{<\,\varepsilon/3} +\underbrace{\sum_{k=K}^{n} \underbrace{\frac{n!}{(n-k)!n^k}}_{\leq 1}\frac{|x|^{k}}{k!}}_{<\,\varepsilon/3} +\underbrace{\sum_{k=K}^{\infty}\frac{|x|^{k}}{k!}}_{<\,\varepsilon/3} < \varepsilon.\]
\(\Box\)
The following criterion of Raabe refines the quotient criterion.
Theorem 7 (Raabe criterion). Let \((a_k)_{k\in\mathbb{N}}\) be a sequence in \(\mathbb{K}\).
If there is a \(k_0\in\mathbb{N}\) and a \(\beta\in\;(1,\infty)\) such that \(a_k\neq 0\) and \[\frac{|a_{k+1}|}{|a_k|} \leq 1-\frac{\beta}{k}\] for all \(k\geq k_0\), then the series \(\sum_{k=1}^{\infty}a_k\) converges absolutely.
If \(\mathbb{K}=\mathbb{R}\) and if there is a \(k_0\in\mathbb{N}\) such that \(a_k\neq 0\) and \[\frac{a_{k+1}}{a_k} \geq 1-\frac{1}{k}\] for all \(k\geq k_0\), then the series \(\sum_{k=1}^{\infty}a_k\) diverges.
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