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.
Discover Bridges
Study Root Criterion
Theorem 1 (Root criterion). Let \(n_0\in\mathbb{N}\) and let \(\sum_{k=1}^\infty a_k\) be a series in \(\mathbb{K}\) and assume that there exists some \(q\in(0,1)\) such that for all \(k\geq n_0\) holds \[\sqrt[k]{|a_{k}|} \leq q.\] Then \(\sum_{k=1}^\infty a_k\) converges absolutely.
Proof: Taking the \(k\)-th power of the inequality \(\sqrt[k]{|a_{k}|}<q\), we obtain that for all \(k\geq n_0\) holds \[|a_k|< q^{k}\] Therefore, the convergent geometric series \(\sum_{k=1}^\infty q^{k}\) is a majorant of \(\sum_{k=1}^\infty a_k\) and thus, we have absolute convergence.\(\Box\)
Theorem 2 (Root criterion (limit form)). Let \(\sum_{k=1}^\infty a_k\) be a series in \(\mathbb{K}\) and assume that \[\limsup_{k \rightarrow \infty} \sqrt[k]{|a_{k}|}<1.\] Then \(\sum_{k=1}^\infty a_k\) converges absolutely.
Example 3. Consider the series \[\sum_{k=1}^\infty \frac{k^5}{3^k}.\] Then we have \[\limsup_{k \rightarrow \infty} \sqrt[k]{\left|\frac{k^5}{3^k}\right|} =\limsup_{k \rightarrow \infty} \frac{\sqrt[k]{k}^5}{3}\] Since we know that \(\sqrt[k]{k}\) converges to \(1\), the whole expression converges to \(\frac13<1\). Hence, the series converges.
Discuss your questions by typing below.
Solve the WeBWorK Exercise
The data for the interactive network on this webpage was generated with pntfx Copyright Fabian Gabel and Julian Großmann. pntfx is licensed under the MIT license. Visualization of the network uses the open-source graph theory library Cytoscape.js licensed under the MIT license.