Advanced Calculus Single Variable

5.3.2 Ratio And Root Tests

A favorite test for convergence is the ratio test. This is discussed next. It is at the other extreme from the alternating series test, being completely oblivious to any sort of cancellation. It only gives absolute convergence or spectacular divergence.

Proof: Suppose r < 1. Then there exists n₁ such that if n ≥ n₁, then

| | 0 < ||an+1||< R | an |

where r < R < 1. Then

|an+1| < R |an|

for all such n. Therefore,

2 p |an1+p| < R |an1+p−1| < R |an1+p−2| < ⋅⋅⋅ < R |an1| (5.7)

(5.7)

and so if m > n, then

< R^m−n₁. By the comparison test and the theorem on geometric series, ∑ converges. This proves the convergence part of the theorem.

To verify the divergence part, note that if r > 1, then 5.7 can be turned around for some R > 1. Showing lim_n→∞

= ∞. Since the n^th term fails to converge to 0, it follows the series diverges.

To see the test fails if r = 1, consider ∑ n⁻¹ and ∑ n⁻². The first series diverges while the second one converges but in both cases, r = 1. (Be sure to check this last claim.) ■

The ratio test is very useful for many different examples but it is somewhat unsatisfactory mathematically. One reason for this is the assumption that a_n > 0, necessitated by the need to divide by a_n, and the other reason is the possibility that the limit might not exist. The next test, called the root test removes both of these objections. Before presenting this test, it is necessary to first prove the existence of the p^th root of any positive number. This was shown earlier in Theorem 2.11.2 but the following lemma gives an easier treatment of this issue based on theorems about sequences.

Proof: Consider the function f

≡ x^p − α.  Then there exists b₁ such that f > 0 and a₁ such that f < 0. (Why?) Now cut the interval into two closed intervals of equal length. Let be one of these which has ff ≤ 0. Now do for the same thing which was done to get from . Continue this way obtaining a sequence of nested intervals with the property that

bk − ak = 21−k(b1 − a1).

By the nested interval theorem, there exists a unique point x in all these intervals. Generalizing Theorem 4.4.6 slightly to include the product of of the terms of finitely many sequences, it follows from Theorem 4.4.11 that

f (x)f (x) = lk→im∞ f (ak)f (bk) ≤ 0

Hence f

= 0. ■

Proof: Suppose first that

^1∕n < R < 1 for all n sufficiently large. Say this holds for all n ≥ n_R. Then for such n,

∘n---- |an| < R.

Therefore, for such n,

n |an| ≤ R

and so the comparison test with a geometric series applies and gives absolute convergence as claimed.

Next suppose

^1∕n ≥ 1 for infinitely many values of n. Then for those values of n, ≥ 1 and so the series fails to converge by the n^th term test. ■

Stated more succinctly the condition for the root test is this: Let

1∕n r = lim nsu→p∞ |an|

then

( ∑∞ { converges absolutely if r < 1 ak test fails if r = 1 k=m ( diverges if r > 1

To see the test fails when r = 1, consider the same example given above, ∑ _n

and∑ _n.

A special case occurs when the limit exists.

Proof: The first and last alternatives follow from Theorem 5.3.9. To see the test fails if r = 1, consider the two series ∑ _n=1^∞

and ∑ _n=1^∞ both of which have r = 1 but having different convergence properties. The first diverges and the second converges. ■