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Integral Calculus

Course: Integral Calculus>Unit 5

Lesson 7: Alternating series test

Alternating series test

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.10 (EK)
When a series alternates (plus, minus, plus, minus,...) there's a fairly simple way to determine whether it converges or diverges: see if the terms of the series approach 0.

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• From to - what is the second condition given for?
Isn't it clear, that if lim(b_n) = 0 when n -> infinity, then {b_n} should be decreasing sequence only? How this sequence could be not decreasing if the limit of b_n (when n -> infinity) is equal zero?
• I can't really explain how any of this works, but I can give a counter-example: sin(pi*n/4)/n is not a decreasing series, but still tends to zero as n approaches infinity.
• The restrictions on this test seem redundant.

from to , Sal gives three restrictions on the series:

1) Bn ≥ 0 for all relevant n (namely positive integers n).
2) lim as Bn→∞ = 0
3) {Bn} is a decreasing sequence.

Don't the first two rules imply the third?
• Consider the function f(n) = x*e^(-n). This will be our Bn.

What do you get at n= 0, n=1, n=2?
f(0) = 0,
f(1) = e^-1 = 0.37,
f(2) = 2/e^-2 = 0.27
So the first 3 terms of the sequence are: [0, 0.37, 0.27].
Notice the b2> b1 this function is not always decreasing! Yet all terms are greater than zero. And the limit as n→∞ = 0. Some functions like this rise a little bit, then fall back down as they go on. So the third rule is necessary.
• Does this test work for an increasing sequence?
• This test is used to determine if a series is converging. A series is the sum of the terms of a sequence (or perhaps more appropriately the limit of the partial sums).

This test is not applicable to a sequence.

Also, to use this test, the terms of the underlying sequence need to be alternating (moving from positive to negative to positive and so on). Therefore the sequence would not be increasing but rather 'oscillating' between positive and negative terms.
• At -: Why is it necessary that the alternating sign be expressed by either (-1)^n or (-1)^(n+1). Aren't there a whole slew of other ways to produced an alternating sign? What about (-1)^(n+2), (-1)^(n^2), or any polynomial exponent with odd integer coefficients?
• We are only talking about the form the series takes on. We know that it alternates, so the question is, is a negative term first, or a positive term. Given n goes from 1 to infinity, the first term of the (-1)^n series will be negative, and the first term of the (-1)^(n+1) series will be positive. That is all that is meant by the form of the series. Why make it any more complicated? It is an alternating series, either the first term is positive, or the first term is negative.
• At He says that we can use other techniques like the limit comparison test, I tired using the limit comparison test but I can't figure out another functions that behaves like this function and that I can figure out It converges ... can someone please tell me how :)
• I don't think we can use the limit comparison test in this case, since the limit comparison require terms in both sequences greater that 0
Maybe he's saying the ratio test, the ratio test also involve limit, and can prove the series converges
• While attempting some practice problems, I couldn't get the correct answer, and this came up as a hint. "This series meets all the conditions for the alternating series test and hence it converges. However, since we can show that ∑n=1∞ n+1n2 diverges by using a comparison test with ∑n=1∞1n. Thus the series converges conditionally." I do not understand what this means, What are the "conditions"?
• I had a similar experience, I'm not sure the hints are properly done on some of these later topics..
• Is this test also called the Leibnitz test for alternating series??
• Yes it is the same! Some books/profs call it alternating and some call it the Leibnitz test.
(1 vote)
• Can you use the ratio test on an alternating series?
• So I just tried the first practice problem in the "Alternating Series" challenge, and I found that one of the hints contradicts what Sal says in this video. At , Sal says, "Now once again, if something does not pass the Alternating Series Test, that does not necessarily mean that it diverges; it just means that you couldn't use the Alternating Series Test to prove that it converges."

The problem hint, however, says, "...Since lim n→∞ an = 1 ≠ 0, the series fails the Alternating Series Test and therefore diverges."

Help? Can the Alternating Series Test be used to prove DIVERGENCE, or only conclusively prove CONVERGENCE?
• The alternating series test only proves an alternating series converges and nothing about whether the series could/will diverge.

Was that the first hint given?
Do you have the URL of the question?
(1 vote)
• How can one use the limit comparison test in this example? I found that it is inconclusive to use 1/n as a comparison, but I don’t know what else to try.