What is the root test?

Question

Elijah Abram · Accepted Answer

Given:

#sum_(n=0)^oo a_n# with #a_n >=0#

and:

#lim_(n->oo) root(n)(a_n) = L#

then:

#0 <=L < 1 => sum_(n=0)^oo a_n# is convergent

# L > 1 => sum_(n=0)^oo a_n =oo# The root test states that given a series with positive terms: #sum_(n=0)^oo a_n# with #a_n >=0# if the succession #{root(n)(a_n)}# is convergent: #lim_(n->oo) root(n)(a_n) = L# then we have: #0 <=L < 1 => sum_(n=0)^oo a_n# is convergent # L > 1 => sum_(n=0)^oo a_n =oo# If #L = 1# then the test does not give us any information. In fact, suppose that: #lim_(n->oo) root(n)(a_n) = L < 1# this means that for any #epsilon > 0# we can find #N# such that: #root(n)(a_n) < L+epsilon# for #n > N# As # L < 1# we can choose #epsilon# such that: #L+epsilon < 1# Then we have, for #n > N#: #root(n)(a_n) < L + epsilon < 1# and elevating both sides to the #n#-th power, which preserves the direction of the inequality: #a_n < (L+epsilon)^n# Now: #sum_(n=0)^oo (L+epsilon)^n# is a geometric series of ratio #L+epsilon < 1# and is absolutely convergent, so also: #sum_(n=0)^oo a_n# is convergent by direct comparison. In the same way if #L > 1# we can establish the inequality: #a_n > (L - epsilon)^n# with #L-epsilon > 1# and determine that #sum_(n=0)^oo a_n# is divergent by direct comparison with a divergent geometric series.

Emily Ammerman · Answer

undefined The root test is a method used in calculus to determine the convergence or divergence of an infinite series. It states that if $ \lim_{n \to \infty} \sqrt[n]{|a_n|} $ exists and is less than 1, then the series $ \sum_{n=1}^{\infty} a_n $ converges absolutely. If the limit is greater than 1 or does not exist, the series diverges. If the limit equals 1, the root test is inconclusive.