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Material Type: Notes; Class: Calculus II; Subject: Mathematics; University: Colgate University; Term: Unknown 1989;
Typology: Study notes
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Proofs of Convergence Tests for Series of Positive Terms
Suppose we have a series
n=1 an^ when^ an^ >^ 0, and we want to know whether this series converges. Because the partial sums form an increasing sequence, the only way the series can fail to converge is by diverging to ∞; so we are really asking whether the partial sums are bounded above. Here are several tests for answering this question:
Integral Test: If f (x) is a decreasing positive function from [1, ∞) to [0, ∞), then the series
f (n) converges if and only if the improper integral
0 f^ (x)^ dx^ converges (i.e., is finite). The (graphical) proof of this appears in a different supplement.
Comparison Test: If 0 < an ≤ bn and
bn converges, then
an also converges.
This is similar to the Comparison Test for improper integrals. The idea is simply that, because the partial sums of
bn are bounded above, then so are the smaller partial sums of
an.
∑ Notes: (1) This statement is logically equivalent to its “contrapositive” statement: If 0^ < an^ ≤^ bn^ and an diverges, then
bn also diverges. (2) This test [respectively, the next test] is used as follows: Given a series
∑ an, pick a series bn about which it is known whether it converges or diverges, and check that an < bn [respectively, that limn→∞(an/bn) is finite and nonzero].
Limit Comparison Test:∑ If 0 < an, bn and limn→∞(an/bn) = L, where L is neither 0 nor ∞, then either an and
bn both converge or both series diverge.
This is a mildly tricky application of the Comparison Test: Pick a number a bit larger than L — L + 1 will do. Because the quotient has limit L, for all n at least as large as some positive integer N we have an/bn < L + 1, and so an < (L + 1)bn. Now the series
(L + 1)bn converges (to L + 1 times whatever the series
bn converges to), so by the Comparison Test (applied only to the terms starting with the N -th one, but that doesn’t affect convergence), we see that
an also converges.
Note: The proof of this test shows that more is true. (It would just make the statement more complicated to try to include it.) For example, if lim(an/bn) = 0, then bn is eventually larger than an; so if
bn converges, then so does
an. And if lim(an/bn) = ∞, then an is eventually larger than bn, so if
bn diverges, then so does an. But for neither of these statements is the converse true.
Ratio Test: If 0 < an and limn→∞(an+1/an) < 1, then
an converges. If the limit is greater than 1, then the series diverges. And if the limit is equal to 1, then the test gives no information.
This application of the Comparison Test is trickier still: Suppose the limit is less than 1, and pick a number r larger than the limit but still less than 1 (for instance, the average of the limit and 1). Then because the limit is less than r for all n at least as large as some positive integer N we have an+1/an < r, i.e., an+1 < anr. So for all k ≥ 1 we have
aN +k < aN +k− 1 r < (aN +k− 2 r)r = aN +k− 2 r^2 < (aN +k− 3 r)r^2 = aN +k− 3 r^3 <... < aN rk^.
So starting with the N -th term, the terms of the series
an are bounded above by the terms of the geometric series
aN rk^ (where the last sum runs over k, not n). This geometric series has common ratio r < 1, so it converges; and hence, so does
an. Suppose the initial limit were greater than 1. Then in a similar way we can see that the terms an are eventually bounded below by the terms of a a geometric series with common ratio greater than 1, so
an diverges.
Note: This test is easy to use, but it often gives a limit of 1, so it is often inconclusive. For example, the Integral Test shows that the p-series
1 /np^ converges for p > 1 and diverges for p ≤ 1, but the Ratio
Test gives a limit of 1 in both cases.
Root Test: If 0 < an and limn→∞ n
an < 1, then
an converges. If the limit is greater than 1, then the series diverges. And if the limit is equal to 1, then the test gives no information.
Suppose the limit is less than 1, and pick a number r between the limit and 1. Then for all “sufficiently large” n we have n
an < r, i.e., an < rn. Thus, for all sufficiently large n the terms in the given series are bounded above by the terms of the convergent geometric series
rn, so the given series also converges. On the other hand, if the limit is greater than 1 and we choose an r between the limit and 1, then the terms of the given series are eventually greater than the terms in the divergent geometric series
rn, so the given series diverges.
Note: The Root Test is harder to apply than the Ratio Test, but it can sometimes show convergence or divergence when the Ratio Test is inconclusive. Like the Ratio Test, however, it is inconclusive on p-series (and hence also on the many series that could be decided with the Limit Comparison Test applied with p-series).