Comparison Test
If a series converges can be checked with different tests.
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Study Comparison Test
The following criterion can be seen as a “series version” of the comparison criterion for sequences.
Theorem 1 (Majorant criterion). Let \(\sum_{k=1}^\infty a_k\) be a series in \(\mathbb{K}\). Moreover, let \(n_0\in \mathbb{N}\) and let \(\sum_{k=1}^\infty b_k\) be a real convergent series such that \(|a_k|\leq b_k\) for all \(k\geq n_0\). Then \(\sum_{k=1}^\infty a_k\) converges absolutely.
Proof: Let \(\varepsilon>0\). By the Cauchy criterion applied to \(\sum_{k=1}^\infty b_k\) there is an \(N\geq n_0\) such that for all \(n\geq m\geq N\) holds \[0\leq \sum_{k=m}^n|a_k|\leq\sum_{k=m}^nb_k=\left|\sum_{k=m}^nb_k\right|<\varepsilon.\] The Cauchy criterion now implies that \(\sum_{k=1}^\infty |a_k|\) converges.\(\Box\)
Remark 2. The series \(\sum_{k=1}^\infty b_k\) with the properties as stated in Theorem about the majorant criterion is called a majorant of \(\sum_{k=1}^\infty a_k\).
Now we present a kind of reversed majorant criterion that gives us a sufficient criterion for divergence.
Theorem 3 (Minorant criterion). Let \(\sum_{k=1}^\infty a_k\) be a real series. Moreover, let \(n_0\in \mathbb{N}\) and \(\sum_{k=1}^\infty b_k\) be a divergent series such that \(a_k\geq b_k\geq 0\) for all \(k\geq n_0\). Then \(\sum_{k=1}^\infty a_k\) diverges.
Proof: We prove the result by contradiction: Let \(\sum_{k=1}^\infty b_k\) be divergent. Assume that \(\sum_{k=1}^\infty a_k\) converges. Then, due to \(a_k\geq b_k\geq 0\), the majorant criterion implies the convergence of \(\sum_{k=1}^\infty b_k\), too. This is a contradiction to our assumption.\(\Box\)
Remark 4. The series \(\sum_{k=1}^\infty b_k\) with the properties as stated in the minorant criterion is called a minorant of \(\sum_{k=1}^\infty a_k\).
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