Sequences Convergence Flowchart
Joseph Breen
Background
First, some context. Everything you’ve studied in calculus up until now has dealt with functions and
integration of these functions. In comparison, sequences and series may initially seem very different.
Because of this, this whole topic can seem kind of intimidating. However, it turns out that everything is the
same! We’re really not doing anything all that different. Here’s why:
The main object of study in math so far has been that of a function. In integral calculus, we take
functions and integrate them. One big question we always ask about a function f is the following: what
happens as x → ∞? Similarly, given an integral of a function (in particular, an improper integral where one
of the bounds goes off to infinity), the question we always ask is the following: does the integral converge or
diverge? This is summarized below:
Functions
Integrals
R∞
f (x)
1
R∞
limx→∞ f (x) = ?
1
f (x) dx
f (x) dx < ∞ ?
Now, suppose I give you a function f , and only evaluate it at x = 1, 2, 3, . . . . This is a sequence!
Let’s call this sequence an . Explicitly, {an } = {f (1), f (2), f (3), . . . }. Remember the main question about
functions from above? We’re going to ask the same question about our sequence: what happens as n → ∞?
In other words, does the sequence converge?
Next, one (informal) way to think about integrating a function is that it is the process of adding up all
the values of the function. If we apply this to our sequence (which again, is just a function), we get a series!
Just as before, we ask the same question about a series as we do with an improper integral: does it converge
or diverge? This is summarized here:
Sequences
Series
P∞
an
1
P∞
limn→∞ an = ?
1
an
an < ∞ ?
So there is a very strong correspondence between functions and sequences, and very strong correspondence between integrals and series. In fact, it’s more than just a correspondence: a sequence is a
1
special type of function, and a series is a special type of integral. So you can apply all the ideas you know
about functions and integrals to sequences and series. This connection makes things like the integral test
seem very obvious, and brings to light why we have very similar tests for both improper integrals and series
(direct comparison, p-test, etc.).
Testing for Convergence
Unfortunately (or fortunately), there are a lot of other tests for series that we can use that we didn’t use with
integrals, so it can be difficult to figure out which one to use. Describing a step-by-step process is difficult
to do in general, but here’s my attempt at describing the process my mind goes through when confronted
P
with a series. Given a series
an , I proceed (roughly) in this order:
1. Divergence Test: If the sequence an doesn’t approach 0, then the series will definitely diverge. If
the sequence does approach 0, we can’t conclude anything, so move on.
2. Special Series: Check to see if the series is in a special form:
• Geometric Series: Looks (roughly) like
P
arn ; you can sometimes take series like
P
2n+1 31−n
term
and morph them into geometric series. If |r| < 1, converges to first1−r
P
• Telescoping Series: Looks (roughly) like
an − bn , because then things cancel out. Find the
formula for the kth partial sum and let k → ∞.
P 1
• P-Series: Looks like
np . If p > 1 converges. If p ≤ 1, diverges.
P
P
• Alternating Series: Looks like (−1)n bn . If bn → 0 and bn is decreasing, (−1)n bn converges.
3. Limit Ratio Test: If the series has a factorial, i.e., something like n!, chances are you should use the
P
|
< 1,
an converges.
limit ratio test1 . If limn→∞ |a|an+1
n|
4. Comparison: If it isn’t a special form and doesn’t have a factorial, maybe we can compare it to
something:
P
P
n|
• Limit Comparison: Take
an and compare it to a simpler series
bn . If 0 < limn→∞ |a
|bn | <
P
P
∞,
an and
bn either both converge or both diverge.
P
P
P
P
• Direct Comparison: If
bn converges and
an ≤ bn , then
an converges. Alternatively,
P
P
P
P
if
bn diverges and
an ≥ bn , then
an diverges.
5. Integral Test: If nothing else works, try using the correspondence from above and look at the integral
version of the series. Given that the function is decreasing, continuous, and positive, the series and
integral will behave the same.
1 The
reason for this is that
(n+1)!
n!
= n + 1, so the terms have a habit of canceling out nicely.
2