Question 4.T.8: (Root Test) Given a series ∑xn, let r = lim sup ^n√|xn|. The...
(Root Test)
Given a series \sum{x_{n}} , let
r = \lim \sup \sqrt[n]{|xn|}.Then
(i) \sum{x_{n}} is absolutely convergent if r < 1.
(ii) \sum{x_{n}} is divergent if r > 1.
(iii) The test is inconclusive if r = 1.
(i) If r < 1, choose a positive number c ∈ (r, 1). Let ε = c − r. By Theorem 3.16(i), there is a positive integer N such that
n ≥ N ⇒ \sqrt[n]{|xn|} < r + ε = c
⇒ |xn| < c^{n}.
Since the geometric series \sum{c^{n}} converges, \sum|{x_{n}}| by comparison, and therefore \sum{x_{n}} is absolutely convergent.
(ii) By Theorem 3.16(iv), there is a subsequence (x_{n_{k}}) of (x_{n}) such that
\sqrt[n_{k}]{|x_{n_{k}}|} → r.
Since r > 1 there is a positive integer N such that
k ≥ N ⇒ \sqrt[n_{k}]{|x_{n_{k}}|} > 1
⇒ |x_{n_{k}}| > 1,
and we conclude that the condition x_{n} → 0, which is a necessary condition for convergence, is not satisfied.
(iii) In Example 4.3 we found that \sum{1/n} was divergent, and in Example 4.7 we established the convergence of \sum{1/n^{2}}. But in both cases r = \lim \sqrt[n]{1/n^{p}} = 1.