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

Click on an **arrow** to get a description of the connection!

Click on an **arrow** to get a description of the connection!

__Show requirements__

__Show consequences__

Concept |
Content |

Power Series |
A sequence of partial sums of polynomial functions. |

**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.