Proofs of Convergence Tests for Series of Positive Terms
Suppose we have a series P∞
n=1 anwhen 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 Pf(n) converges
if and only if the improper integral R∞
0f(x)dx converges (i.e., is finite).
The (graphical) proof of this appears in a different supplement.
Comparison Test: If 0 < an≤bnand Pbnconverges, then Panalso converges.
This is similar to the Comparison Test for improper integrals. The idea is simply that, because the
partial sums of Pbnare bounded above, then so are the smaller partial sums of Pan.
Notes: (1) This statement is logically equivalent to its “contrapositive” statement: If 0 < an≤bnand
Pandiverges, then Pbnalso diverges.
(2) This test [respectively, the next test] is used as follows: Given a series Pan, pick a series
Pbnabout 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, bnand limn→∞(an/bn) = L, where Lis neither 0 nor ∞, then either
Panand Pbnboth 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 nat least as large as some positive integer Nwe have
an/bn< L + 1, and so an<(L+ 1)bn. Now the series P(L+ 1)bnconverges (to L+ 1 times whatever the
series Pbnconverges 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 Panalso 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 bnis eventually larger than an; so if Pbnconverges,
then so does Pan. And if lim(an/bn) = ∞, then anis eventually larger than bn, so if Pbndiverges, then
so does an. But for neither of these statements is the converse true.
Ratio Test: If 0 < anand limn→∞(an+1/an)<1, then Panconverges. 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 rlarger than the limit but still less than 1 (for instance, the average of the limit and 1). Then
because the limit is less than rfor all nat least as large as some positive integer Nwe have an+1/an< r,
i.e., an+1 < anr. So for all k≥1 we have
aN+k< aN+k−1r < (aN+k−2r)r=aN+k−2r2<(aN+k−3r)r2=aN+k−3r3< . . . < aNrk.
So starting with the N-th term, the terms of the series Panare bounded above by the terms of the geometric
series PaNrk(where the last sum runs over k, not n). This geometric series has common ratio r < 1, so it
converges; and hence, so does Pan.
Suppose the initial limit were greater than 1. Then in a similar way we can see that the terms anare
eventually bounded below by the terms of a a geometric series with common ratio greater than 1, so Pan
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 P1/npconverges for p > 1 and diverges for p≤1, but the Ratio
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