If a series converges can be checked with different tests.

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Concept |
Content |

Quotient Criterion |
An important criterion to prove absolute convergence by means of ratios of the underlying sequence's terms. |

Root Criterion |
An important criterion to prove absolute convergence by means of the behavior of the n-th roots of the underlying sequence's terms. |

Integral Comparison Test |
Compare improper Riemann integrals to infinite series. |

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.\(\Box\)

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.\(\Box\)

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