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A GUIDE TO THE LIMIT COMPARISON TEST
Joseph Breen The limit comparison test is the GOAT infinite series convergence test, but knowing when and how to use it effectively can be difficult. This guide explains the intuition, subtleties, and heuristics of the test and hopefully provides enough elucidating examples. The blue links below are hyperlinks to each section.
Contents
1 The statement of the limit comparison test 1
2 When and how to use limit comparison 2 2.1 A quick summary............................................ 2 2.2 Case I: Isolating obvious dominant behavior............................ 2 2.3 Case II: Pesky logarithms........................................ 3 2.4 Case III: Trig functions and other oddities (Taylor polynomials)................. 4 2.5 Case IV: A last ditch resort....................................... 6
3 Some examples 7
1 The statement of the limit comparison test
In order to use limit comparison, we have to know the statement. I’ll provide the mathematical statement, but also how you should think about the statement.
Theorem 1.1 (Limit comparison test.). Let
n=1 an^ be an infinite series with^ an^ >^0. Let^ bn^ >^0 be a positive sequence.
(i) If limn→∞ a bnn is a finite, positive number, then
n=1 an^ and^
n=1 bn^ either both converge or diverge.
(ii) If limn→∞ a bnn = 0, then if
n=1 bn^ converges, then^
n=1 an^ also converges. (iii) If limn→∞ a bnn = ∞, then if
n=1 bn^ diverges, then^
n=1 an^ also diverges.
LCT only works with positive series
an ≤
bn
an ≥
bn
an '
bn
WARNING : One thing to keep in mind about cases (ii) and (iii) above is that you need to be careful if you do limit comparison and you get a limit of 0 or ∞. In particular, let’s say you pick a series
bn to compare with, and you know that
bn diverges. If you compute that limn→∞ a bnn = 0, you have argued that
an ≤
bn. Since
bn diverges, this doesn’t tell you anything! Similarly, if you pick a series
bn to compare with, you that
bn converges, and you compute that limn→∞ a bnn = ∞, you have argued that ∑ an ≥
bn and this doesn’t tell you anything!
2 When and how to use limit comparison
In my mind, there are three main scenarios in which limit comparison is helpful, and then a fourth catch- all scenario. The first subsection gives a summary, and the following subsections below describe these scenarios in detail.
2.1 A quick summary
If you are given a series...
(I) ...of the form ∑∞ n=
[dominant term] + [fluff] [dominant term] + [fluff] then run limit comparison against the series ∑^ ∞ n=
[dominant term] [dominant term]
Typically this will be a p -series. Also, if you do this correctly your limit computation should give you limn→∞ a bnn = 1 and thus you are in scenario (i) of the limit comparison test.
(II) ...with a ln n that you can’t ignore , then generate a bn to compare with by replacing ln n with nsmall #. The small number depends on what else is in the series. When you do this you will almost cer- tainly be in scenario (ii) or (iii) of the limit comparison test , so you need to be careful about your conclusion.
ln n nsmall #
(III) ...with something like sin
n
, generate a bn to compare with by replacing the sin
n
term with (^) n^1. There are other useful approximations (based on Taylor polynomials) but in 31B, sin
n
is the mostly likely suspect.
sin
n
≈ (^1) n
(IV) ...and none of the above situations above apply. If you think that the series diverges, just run limit comparison against (^1) n. If you think the series converges, run limit comparison against (^) n^12. If you’re unsure, just try both!
2.2 Case I: Isolating obvious dominant behavior
This is the main situation in which limit comparison applies, and it is also the most straightforward. Given a series that exhibits obvious dominant terms in the numerator and denominator, you can generate a bn to compare with by isolating this dominant behavior. This is best explained with an example: consider the series (^) ∞ ∑ n=
n √^2 + 1 + sin n n^7 + n^5 + 1
My first thought when faced with the above series is that there is a lot of fluff that I can ignore. For example, in the numerator, there are three different terms: n^2 , 1 , and sin n. As n grows to infinity, the most important term is n^2. (The 1 term does not grow at all, and sin(n) dances around in the interval [− 1 , 1]. Note that we are not in Case III described above because sin(n) is much different than sin
n
.) Likewise, in the denominator
This warning shows that even though logarithmic growth is insanely slow, you cannot simply ignore it. You do have to budget a small amount of growth to absorb the logarithm. This is where the heuristic
ln n nsmall^ #
comes into play. If we run this heuristic on our summand, we get
ln n n^2
nsmall^ # n^2
n^2 −(small^ #)^
Because we want to show that our series converges, we want to make sure that the series
∑^ ∞
n=
n^2 −(small^ #)
converges. This means that we need 2 − (small #) > 1 and so 0 < small # < 1. In this case, we can pick small # = 12 and thus run limit comparison against (^) n^1 32. Indeed,
n^ lim→∞
ln n n^2 1 n 32
= (^) nlim→∞ ln n n 12
since ln n na ∑ (^) ln n n 32
n 32 Because we got a limit of 0 and we know that
n= 1 n 32 converges, we can successfully apply scenario (ii) of the limit comparison test to conclude that
n= ln n n^2 also converges.
2.4 Case III: Trig functions and other oddities (Taylor polynomials)
The main take away from this case is the following approximation:
sin
n
n.
Where does this come from, and are there other useful approximations?
The first approximation: sin
n
≈ 1 n
Let’s talk about the function sin x first, and then I will describe the more general situation. Consider the graph of y = sin x. The tangent line at x = 0 is y = x, because sin(0) = 0 and cos(0) = 1. We know that close to 0 , this tangent line should be a good approximation to the function. Indeed, we can graph both near 0 to confirm this:
x
y y = sin x y = x
It is clear that sin x ≈ x as long as x is close to 0. The fraction (^1) n is close to 0 for large values of n, so indeed sin
n
≈ (^1) n. We can exploit this fact to generate useful candidates for limit comparison. For example, consider the series (^) ∞ ∑ n=
sin
n
n
Based on the above discussion, my intuition tells me that
∑^ ∞ n=
sin
n
n
∑^ ∞
n=
1 n n
∑^ ∞
n=
n^2
As usual, the formalization of this intuition is limit comparison. Let’s compute:
n^ lim→∞
sin( (^) n^1 ) n 1 n^2
= (^) nlim→∞ sin
n
1 n
= (^) xlim→∞ sin
x
1 x (L′H) = (^) xlim→∞
cos
x
· − (^) x^12 − (^) x^12 = cos(0) = 1.
Since 1 is a finite, positive number, we are in scenario (i) of the limit comparison test. This implies that the
series
n=
sin( (^1) n ) n and^
n= 1 n^2 behave the same, just as we suspected. Since the latter is a convergent p-series, the former converges as well!
Other useful approximations
Case III doesn’t end with the approximation of sin
n
. Because sin x ≈ x for x small, we get a host of other similar approximations:
sin
n^2
n^2 sin
n^3
n^3 sin
√^1
n
√^1
n
and so on and so forth. Furthermore, by manipulating these approximations we can go even further:
sin^2
√^1
n
√^1
n
n √ sin
n^3
n^3
n 32
In general, approximations like sin x ≈ x for x small can give you intuition about the behavior of many different series.
WARNING : The approximation sin x ≈ x only holds when x is small. In order to use this approximation, the thing inside of the sine function needs to tend to 0. In particular, if you see something like sin(n) in a series, it would be wildly incorrect to claim that sin(n) ≈ n.
Bigger size
Convergent series
Divergent series
n= 1 n^2
n= 1 n
Most convergent series
Most divergent series
This relative size comparison is just a heuristic and not actually precise, but it is helpful nonetheless. The point is this: if you want to conclude convergence via comparison, you need to compare with some- thing bigger that also converges, and if you want to conclude divergence via comparison, you need to compare with something smaller that also diverges. Because
n= 1 n^2 is heuristically bigger than most convergent series, it’s a decent candidate for a convergent comparison argument. Likewise with
n= 1 n and a divergence comparison argument.
3 Some examples
Here some worked out examples that illustrate the principles in this guide. I encourage you to try them yourself first to test your mastery of limit comparison!
Example 3.1. Determine whether the series ∑^ ∞ n=
n^2 + cos^2 n 1 +
n^5 + 1
converges or diverges.
Solution. This series looks like a Case I infinite series, in that there are clearly dominant terms in both the numerator and denominator surrounded by some unimportant fluff that we can throw away. In particular, we expect n^2 + cos^2 n 1 +
n^5 + 1
n^2 √ n^5
for large n. Thus, we run limit comparison with bn = n √^2 n^5. Note that
n^ lim→∞
n^2 +cos^2 n 1+√n^5 + √^ n^2 n^5
= (^) nlim→∞
1 + cosn^22 n √^1 n^5 +
1 + (^) n^15
This last equality uses the fact that 0 ≤
∣ cos (^2) n n^2
∣ ≤^ n^12 →^0. Since^1 is a finite, positive number, we are in scenario (i) of the limit comparison test. Because the series
∑^ ∞ n=
√^ n^2 n^5
∑^ ∞
n=
n 12
is a divergent p-series, it follows that
n= n^2 +cos^2 n 1+√n^5 +1 diverges as well.
Example 3.2. Determine whether the series ∑^ ∞ n=
1 + n n 74 ln n
converges or diverges.
Solution. Because the denominator is n 74 ln n, there is logarithmic growth that we cannot ignore, which suggests this is a Case II series. The numerator 1 + n has fluff that we can ignore, and in the denominator we will use the heuristic ln n nsmall^ #: 1 + n n 74 ln n
n n 74 nsmall^ #^
n 34 +(small^ #)^
Because we can choose (small #) as small as we like, and the power of n in the denominator is 34 +(small #), we expect this series to diverge (if we pick a sufficiently small number, the exponent will be ≤ 1 ). In particular, we can choose (small #) = 14 and thus run limit comparison against (^) n^1. Let’s compute:
n^ lim→∞
1+n n 74 ln n 1 n
= (^) nlim→∞ 1 + n n 34 ln n
= (^) xlim→∞ 1 + x x 34 ln x (L′H) = (^) xlim→∞
3 4 x − (^14) ln x + x− 14 =^ ∞.
This last equality comes from the fact that ln^ x x 14 → 0 and 1 x 14 → 0 as x → ∞. Because the limit is ∞, we are in scenario (iii) of the limit comparison test. Since the series
n= 1 n diverges, we can conclude that the series
n= 1+n n 74 ln n diverges as well.
Example 3.3. Determine whether the series ∑^ ∞ n=
sin
n 32
ln(n + 2)
converges or diverges.
Solution. Because there is only one term of the summand, and it has both sin
1 n 32
and ln(n + 2), this series
is a mixture of Case II and Case III. We will thus use the heuristics sin x ≈ x for x small and ln n nsmall^ #:
sin
n 32
ln(n + 2)
n 32
· nsmall^ #^ =
n 32 −(small^ #)^
It’s possible to L’Hopital this limit (it would take four applications), but I’ll do something lazier. Make a variable substitution m =
n. Then n^2 = m^4 and as n → ∞, m → ∞. The limit turns into
n^ lim→∞
n^2 e
√n = (^) mlim→∞^ m
4 em^.
Because we already know that ma^ bm^ for any a, b > 0 , it follows that
m^ lim→∞
m^4 em^
Thus, we are in scenario (ii) of the limit comparison test and since we know that
n= 1 n^2 converges, we can conclude that
n=1 (^) e√^1 n converges as well.
Example 3.5. Determine whether the series
∑^ ∞ n=
e √^1 n − 1 n
converges or diverges.
Solution. Because of the e √^1 n term, we can approximate the behavior in the numerator using Taylor polyno- mials. In particular, recall that the degree 1 Taylor polynomial centered at 0 for ex^ is 1 + x. Thus, we expect ex^ − 1 ≈ x for x small and thus e √^1 n − 1 ≈ √^1 n. Thus, we will run limit comparison against
√^1 n n
n 32
Note that
n^ lim→∞
e √^1 n − 1 n 1 √n n
= (^) nlim→∞ e √^1 n − 1 √^1 n
= (^) xlim→∞ e √^1 x − 1 √^1 x
= (^) xlim→∞ e √^1 x = e^0 = 1
by L’Hopital’s rule. Since we got a limit of 1 , we are in scenario (i) of the limit comparison test. Since the
series
n= 1 n 32 converges, the series
n= e √^1 n − 1 n converges as well.
Example 3.6. Determine whether the series ∑^ ∞ n=
2 n^ + ln n en^ + sin n
converges or diverges.
Solution. Note that even though there is a ln n in the numerator , we are not in Case II. Because the ln n is a separate term from 2 n, and 2 n^ grows faster, this instance of ln n counts as fluff. Likewise, the sin n in the denominator does not indicate that we are in Case III or anything like that; it counts as fluff. This is purely a Case I infinite series.
We will run limit comparison against 2 n en^. Note that
n^ lim→∞
2 n+ln n en+sin n 2 n en
= (^) nlim→∞
1 + ln 2 nn 1 + sinen^ n
The limit in the numerator comes from the fact that ln n na^ bn^ and the limit in the denominator is a result of 0 ≤
∣∣ sin n en
≤ (^) e^1 n → 0. Because we got a limit of 1 , we are in scenario (i) of the limit comparison test. Because (^) ∞ ∑ n=
2 n en^
∑^ ∞
n=
e
)n
is a convergent geometric series ( 2 < e), we can conclude that the series
n= 2 n+ln n en+sin n converges as well.
Example 3.7. Determine whether the series ∑^ ∞ n=
ln(ln n) (n^5 + 1) 14
converges or diverges.
Solution. Because of the logarithm mess in the numerator, we are definitely in Case II. Note that even though the term is ln(ln n) instead of just ln n, the same principles. In particular, our heuristics lead to:
ln(ln n) (n^5 + 1) 14
nsmall^ # n 54
n 54 −(small^ #)^
Because we can choose (small #) as small as we like, and the power of n in the denominator is 54 −(small #), we expect this series to converge (we can choose (small #) so small that the exponent remains larger than 1 ). In particular, we can pick small # = 18 and run limit comparison with 1 n 98
. Indeed,
n^ lim→∞
ln(ln n) (n^5 +1) 14 n 18 n 54
= (^) nlim→∞^ ln(ln^ n) n
18 ·^
1 + (^) n^15
) 14 = 0^ ·^
The limit computation uses the fact that ln n na^ and so ln(ln n) ln n na. Since we got a limit of 0 , we are in scenario (ii) of the limit comparison test. Because we know that
∑^ ∞ n=
n (^18)
n
∑^ ∞
n=
n (^98)
converges, we can conclude that
n=
ln(ln n) (n^5 +1) 14 converges as well.
