8.5. Ratio Test, Root Test, and Summary of Tests http://www.ck12.org
8.5 Ratio Test, Root Test, and Summary of Tests
Ratio Test
We have seen the Integral Test, (Limit) Comparison Test and Alternating Series Test which impose conditions on
the sign ofan. We are going to introduce two tests for a stronger version of convergence that do not.
Definition
(Absolute convergence)
A series∑anisabsolutely convergentif the series of the absolute values∑|an|is convergent.
To this end, we need to distinguish the other type of convergence.
Definition
(Conditional convergence)
A series∑anisconditionally convergentif the series is convergent butnotabsolutely convergent.TheoremIf∑anis absolutely convergent, then it is convergent.
The proof is quite straightforward and is left as an exercise. The converse of this theorem is not true. The series
∑(−^1 n)n−^1 is convergent by the Alternating Series Test, but its absolute series,∑^1 n(the harmonic series), is divergent.
Example 1∑cosn 2 nθis absolutely convergent since∣∣cosn 2 nθ∣∣≤n^12 for any 1≤n,θ, and∑n^12 is convergent (e.g. by the
p−test). Indeed, by the Integral and Comparison tests,∑cosnpnθis absolutely convergent for anyθandp>1.
The limit of the ratio
∣∣
∣ana+n^1∣∣
∣gives us a comparison of the tail part (i.e∑nLargean) of the series∑anwith a geometric
series.
Theorem(The Ratio Test) Let∑anbe a series of non-zero numbers∗.
(A) If limn→∞
∣∣
∣ana+n^1∣∣
∣=α<1, then the series is absolutely convergent.(B) If limn→∞
∣∣
∣ana+n^1∣∣
∣=α>1 or limn→∞∣∣
∣ana+n^1∣∣
∣=∞then the series is absolutely divergent.(C) If limn→∞
∣∣
∣ana+n^1∣∣
∣=α=1 then the test is inconclusive.
∗: we could ignore the zero-valuedan’s as far as the sum is concerned.
Proof. (A) The proof is by comparison with a geometric series.
Ifα<1, thenα <α+ 21 <1. It follows from the definition of limit that there is an integerN,
∣∣
∣ana+n^1∣∣
∣<α+ 21 for all
n≥N. Letβ=α+ 21 Then|aN+ 1 |<β|aN|,|aN+ 2 |<β|aN+ 1 |<β^2 |aN|,...and recursively we have|an+ 1 |<βn−N+^1 |aN|
forn≥N, and∑∞n=N|an|<|aN|∑∞n=Nβn−Nwhich is finite. Combining with the finitely many terms,∑∞n= 0 |an|is still
finite.
(B) A similar argument concludes limn→∞an 6 = 0 .So the series is divergent.
Example 2Test the series∑Ann!for absolute convergence whereAis a constant.
Solution. Letan=Ann!. Then
∣∣
∣ana+n^1∣∣
∣=(An+n+ 11 )!·An!n=n+A 1 →0 asn→∞. So by the Ratio Test, the series is absolutely
convergent for any constantA. Indeed, the sum iseAwhich is very large for largeA, but still finite.