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Tips and instructions on how to use various tests to determine the convergence or divergence of mathematical series. Topics covered include the geometric series test, p-series test, integral test, comparison test, limit comparison test, divergence test, and alternating series test. Each test is explained in detail, including how to apply it and when it is most effective.
Typology: Lecture notes
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Tips on Using Tests of Convergence
n=0 ar
n. We say that the series converges if and only if |r| < 1 and the sum is given by,
n=0 ar
n (^) = a 1 −r ,^ where^ a^ is the first term and r is the common ratio. Also note that
∑k n=0 ar
n (^) = a(1−rk+1) 1 −r.^ To understand this formula, look for its proof into your notes. Note that terms of the series could even be negative.
n=
1 np^ converges if and only if^ p >^ 1. This test can be proved using integral test(which is described next). You will probably never get to use this test individually in a problem but nevertheless this test has important role to play as it supports the working of other tests. We generally use this test along with comparison and limit comparison test.
n=1 an converges if and only if
1 f^ (x)^ dx^ converges. Make sure to check all the properties before you apply this test. In general, you use this test when the terms have the configuration of a derivative. For example:
∑ (^) ln n n ,^
nln np^ ,^
ne−n^ and etc. The convergence or the divergence for some of the above mentioned examples may also be shown using other tests, for instance, we could use comparison test for the first one, Root test or the Ratio test for the third one.
an is a series with positive terms and
bn converges then
an also converges.
bn diverges then
an also diverges.
Typically, you choose bn to be either the p-series or the geometric series(basically the series about which you know quite a lot in terms of its convergence). Please be warned about the right inequalities. It is easy to fall into the trap of justifying the wrong inequality.
an, an ≥ 0) only. Again you choose bn as in the case of the comparison test. The only difference is(which makes this test easy to use) that instead of checking for in- equalities, you look for the limit of a bnn.
Formally speaking, this states that if
an is a series with positive terms and you choose
bn such that limn→∞ a bnn is non-zero and finite then
an converges(diverges)
if and only if
bn converges(diverges).
Typically, you use this test when you are a given a series with nth^ term given by a rational function in term of n. This by no means imply that you cannot use this test for series other the ones that are given by rational functions, for example, this test works best for
sin (^) nπ 2. Keep in mind that this test fails if the limit turns out to be zero or infinity. In this case, you should look for other test. It is possible that the same series might converge or diverge using some other test.
an must diverge. The converse of this is not true. Other than checking the divergence of the series, this test also allows us find the limit of sequences. If the series
an converges then limn→∞ an = 0. This is quite useful. For instance, if you recall we found limn→∞ (^) nnn! = 0 using squeeze theorem. Another way to see this, show that the series
∑ (^) n! nn^ converges which can be done using ratio test.
(−1)nan if in addition to lim an = 0 we have that all the terms an are positive and decreasing.
Formally speaking, if
(−1)nan is a series with an ≥ 0 for all n such that
then
(−1)nan converges.
The converse of the above is not true. In other words, the above test is inconclusive if an’s fail to decrease. Note that if lim an 6 = 0 then lim(−1)nan does not exist which implies that
(−1)nan fail to diverge by divergence test.
This test states that given any series
an.
an converges absolutely.
an diverges.
(b) Determine whether the series
n→∞(−1) n(√n + 1 + √n)n (^) converges or diverges.
We do not NEED Root test for this one, though, we can use if we want. Note that limn→∞(
n + 1+
n)n^ == ∞)∞^ = ∞. This implies that limn→∞(−1)n(
√ n^ + 1+ n)n^ does not exist. Hence the given series
n→∞(−1)
n(√n + 1 + √n)n (^) diverges by divergence test.
If the terms are rational functions of n then this test is very difficult to apply and also fails almost every time. Even for simple example such as
n this test fails. Indeed if an = (^1) n and we consider
lim n→∞ |an|^1 /n^ = lim n→∞
n
1 /n = lim n→∞
n^1 /n^
using the fact that limn→∞ n^1 /n^ = 1. This implies that the root test is inconclusive.
Summary of Some Important Tips
Last Remark: Absolute convergence means everything converges, that is,
|an| converges which further implies that
an convergence. Thus it can thought of as the “strong convergence”. On the other hand, conditional convergence means that the series
an converges BUT
|an| diverges. Thus we can think of this convergence as “ weak convergence” or “partial convergence”.