A Guide to the Limit Comparison Test

AGUIDETOTHE 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 test1

2 When and how to use limit comparison2 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 LCT only 7 works with positive series 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.

P∞ Theorem 1.1 (Limit comparison test.). Let n=1 an be an inﬁnite series with an > 0. Let bn > 0 be a positive sequence.

an P∞ P∞ (i) If limn→∞ is a ﬁnite, positive number, then an and bn either both converge or bn n=1 n=1 diverge. P P an P∞ P∞ an ≤ bn (ii) If limn→∞ = 0, then if bn converges, then an also converges. bn n=1 n=1

an P∞ P∞ (iii) If limn→∞ = ∞, then if bn diverges, then an also diverges. bn n=1 n=1

P P P P 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 P you do limit comparison and you get a limit of 0 or ∞. In particular, let’s say you pick a series bn to P an compare with, and you know that bn diverges. If you compute that limn→∞ b = 0, you have argued P P P n P that an ≤ bn. Since bn diverges, this doesn’t tell you anything! Similarly, if you pick a series bn P an to compare with, you that bn converges, and you compute that limn→∞ b = ∞, you have argued that P P n an ≥ bn and this doesn’t tell you anything!

1 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 ∞ X [dominant term] + [ﬂuff] [ ] + [ ] n=1 dominant term ﬂuff then run limit comparison against the series

∞ X [dominant term] . [ ] ln n nsmall # n=1 dominant term

Typically this will be a p-series. Also, if you do this correctly your limit computation should give an you limn→∞ = 1 and thus you are in scenario (i) of the limit comparison test. bn

small # (II) ...with a ln n that you can’t ignore, then generate a bn to compare with by replacing ln n with n . 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. 1 1 sin n ≈ n

1 1 1 (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.

(IV) ...and none of the above situations above apply. If you think that the series diverges, just run limit 1 1 comparison against n . If you think the series converges, run limit comparison against n2 . 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 ∞ X n2 + 1 + sin n √ . 7 5 n=1 n + n + 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: n2, 1, and sin n. As n grows to infinity, the most important term is n2. (The 1 term does not grow at all, and sin(n) dances around in the interval [−1, 1]. Note that we are 1 not in Case III described above because sin(n) is much different than sin n .) Likewise, in the denominator

2 n7 n5 1 there√ are three terms under a square root: , , and . Thus, the dominant behavior in the denominator is n7. All of this leads me to the following intuition:

∞ ∞ ∞ ∞ X n2 + 1 + sin n X n2 X n2 X 1 √ ≈ √ = 7 = 3 . 7 5 7 2 2 n=1 n + n + 1 n=1 n n=1 n n=1 n

∞ ∞ 2 P 1 P n√ +1+sin n Because n=1 3 is a convergent p-series, this intuition suggests that n=1 7 5 should also con- n 2 n +n +1 verge. The limit comparison test is the way to formalize this intuition! Indeed,

2 n√ +1+sin n a 7 5 lim n = lim n +n +1 n2 n→∞ bn n→∞ √ n7 1 + 1 + sin n = lim n2 n2 n→∞ q 1 1 1 + n5 + n7 1 + 0 + 0 = √ 1 + 0 + 0 = 1.

∞ 2 1 P n√ +1+sin n Because is a ﬁnite, positive number, we are in case (i) of the limit comparison test: n=1 n7+n5+1 and P∞ 1 n=1 3 either both converge or both diverge. Because the latter series converges, we have concluded by n 2 ∞ 2 P n√ +1+sin n limit comparison that n=1 n7+n5+1 converges as well!

2.3 Case II: Pesky logarithms

Logarithms introduce some subtleties in limit comparison. The main takeaway from this section is the following fact: ln n na for any a > 0. ln n grows very In fact, slowly (ln n)p na for any a > 0 and any p > 0. All of this means that we can absorb the growth of a logarithm with a very small power of n. In particular, if there is a logarithm that you can’t ignore in a series, you should generate a candidate for comparison by replacing the ln n term with a very small power of n. By doing this, you absorb the logarithmic growth and will likely be in scenario (ii) or (iii) of the limit comparison test. It’s easiest to explain this through an example. Consider the following series:

∞ X ln n . n2 n=1

Because the denominator grows like n2 and the denominator grows much slower than n2, my intuition suggests that this series should converge. Let’s formalize this with limit comparison.

1 WARNING: In a situation like this, you cannot just ignore the ln n and run comparison against n2 . It will fail! Let’s check: ln n ln n 1 2 P ≥ P lim n = lim ln n = ∞. n2 n2 n→∞ 1 n→∞ n2 Go back and read scenario (iii) of the limit comparison test carefully: because we got a limit of ∞, we can only make a conclusion if we know that the corresponding series in the denominator diverges. But P∞ 1 n=1 n2 . So the test is inconclusive!

3 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

small # ln n n

comes into play. If we run this heuristic on our summand, we get ln n nsmall # 1 = . n2 n2 n2−(small #) Because we want to show that our series converges, we want to make sure that the series ∞ X 1 n2−(small #) n=1 converges. This means that we need 2 − (small #) > 1 and so 0 < small # < 1. In this case, we can pick 1 1 small # = 2 and thus run limit comparison against 3 . Indeed, n 2 since ln n na ln n n2 ln n lim = lim 1 = 0. P ln n P 1 n→∞ 1 n→∞ 3 ≤ 3 3 n 2 n 2 n 2 n 2 P∞ 1 Because we got a limit of 0 and we know that n=1 3 converges, we can successfully apply scenario (ii) n 2 P∞ ln n of the limit comparison test to conclude that n=1 n2 also converges.

2.4 Case III: Trig functions and other oddities (Taylor polynomials)

The main take away from this case is the following approximation: 1 1 sin ≈ . n n Where does this come from, and are there other useful approximations?

1 1 The first approximation: sin n ≈ 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:

2 y y = sin x y = x 1

x −2 −1 1 2

−1

−2

4 1 It is clear that sin x ≈ x as long as x is close to 0. The fraction n is close to 0 for large values of n, so indeed 1 1 sin n ≈ n . We can exploit this fact to generate useful candidates for limit comparison. For example, consider the series ∞ 1 X sin n . n n=1 Based on the above discussion, my intuition tells me that

∞ 1 ∞ 1 ∞ X sin X X 1 n ≈ n = . n n n2 n=1 n=1 n=1

As usual, the formalization of this intuition is limit comparison. Let’s compute:

sin( 1 ) n sin 1 sin 1 lim n = lim n = lim x n→∞ 1 n→∞ 1 x→∞ 1 n2 n x cos 1 · − 1 (L0H) = lim x x2 = cos(0) = 1. x→∞ 1 − x2 Since 1 is a ﬁnite, positive number, we are in scenario (i) of the limit comparison test. This implies that the sin 1 P∞ ( n ) P∞ 1 series n=1 n and n=1 n2 behave the same, just as we suspected. Since the latter is a convergent p-series, the former converges as well!

Other useful approximations

1 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:

1 1 sin ≈ n2 n2 1 1 sin ≈ n3 n3 1 1 sin √ ≈ √ n n and so on and so forth. Furthermore, by manipulating these approximations we can go even further:

1 1 2 1 sin2 √ ≈ √ = n n n s 1 r 1 1 sin 3 ≈ 3 = 3 . n n n 2

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.

5 The full story: Taylor polynomial approximations

More generally, we can capture the behavior of other sequences by using knowledge of Taylor polynomials. For example, recall that the degree 2 Taylor polynomial for cos x at 0 is

x2 T (x) = 1 − . 2 2

x2 This means that for x near 0, we have the approximation cos x ≈ 1 − 2 . We can use this approximation just like we used the one for sin x above. In particular, given a series like

∞ X 1 1 − cos n2 n=1

we could identify the general behavior of the summand as

1 1 1 2 1 1 − cos ≈ = . n2 2 n2 2n4

P∞ 1 P∞ 1 We could then use limit comparison to verify that n=1 1 − cos n2 behaves the same as n=1 2n4 , and thus converges. As another quick example of generating such approximations, recall that the degree 1 Taylor polynomial for ex at 0 is

T1(x) = 1 + x.

Thus, we expect ex ≈ 1 + x for x near 0. This leads to sequential approximations like

1 1 1 − e n ≈ − n and so on and so forth. P∞ The general description is this: given a series n=1 f(sn) where f is some function and sn → 0, we can generate a candidate for limit comparison by considering Tk(sn), where Tk(x) is a Taylor polynomial for f centered at 0.

2.5 Case IV: A last ditch resort

If you are given an inﬁnite series and none of the above cases apply (and nothing else seems to work), 1 1 sometimes you can just run limit comparison against either n or n2 . This won’t always work, but many times it will. Explicitly:

P∞ 1 - If you think the series n=1 an diverges, run limit comparison against n . P∞ 1 - If you think the series n=1 an converges, run limit comparison against n2 . P∞ 1 The reason why this is an effective catch-all strategy is that in the world of convergent series, n=1 n2 is P∞ 1 one of the "biggest" inﬁnite series, i.e., it converges pretty slowly. Likewise, n=1 n is one of the smallest divergent series, in that it diverges very slowly. Here’s what I mean in a picture:

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n Bigger size lim →∞ n 1+

2 Convergent series Divergent series +cos √ √ n n n 2 5 5 +1 2 n lim = P P n →∞ n ∞ + 1 n ∞ n =1 =1 n X 2 ∞ =1 cos + n √ √ n 1 1 2 b 1 n n + 1 n n 5 + 1 5 2 = + 1 + 2 cos + 7 √ n Because . √ q cos n n n 2 n + 1 5 ≈ 5 2 2 oethat Note . 1 + 2 n √ n n n n otcnegn series convergent Most series divergent Most 1 2 5 5 = P + 0 n ∞ =1 0 + 1 √ n 1 2 0 + 1 shuitclybge hnmost than bigger heuristically is 1 = . P n ∞ =1 n 1 2 0 ≤ cos n ≤ 1 → 0 1 This last equality uses the fact that n2 n2 . Since is a ﬁnite, positive number, we are in scenario (i) of the limit comparison test. Because the series

∞ ∞ X n2 X 1 √ = 1 5 2 n=1 n n=1 n

∞ 2 2 p P n +cos√ n is a divergent -series, it follows that n=1 1+ n5+1 diverges as well.

Example 3.2. Determine whether the series

∞ X 1 + n 7 4 n=1 n ln n

converges or diverges.

7 Solution. Because the denominator is n 4 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 1 = . 7 7 3 +( #) n 4 ln n n 4 nsmall # n 4 small 3 Because we can choose (small #) as small as we like, and the power of n in the denominator is 4 +(small #), we expect this series to diverge (if we pick a sufﬁciently small number, the exponent will be ≤ 1). In 1 1 particular, we can choose (small #) = 4 and thus run limit comparison against n . Let’s compute:

1+n 7 1 + n 1 + x lim n 4 ln n = lim = lim n→∞ 1 n→∞ 3 x→∞ 3 n n 4 ln n x 4 ln x 0 1 (L H) = lim 1 1 = ∞. x→∞ 3 − 4 − 4 4 x ln x + x ln x 1 This last equality comes from the fact that 1 → 0 and 1 → 0 as x → ∞. x 4 x 4 P∞ 1 Because the limit is ∞, we are in scenario (iii) of the limit comparison test. Since the series n=1 n P∞ 1+n diverges, we can conclude that the series n=1 7 diverges as well. n 4 ln n

Example 3.3. Determine whether the series

∞ X 1 sin 3 ln(n + 2) 2 n=1 n

converges or diverges.

1 Solution. Because there is only one term of the summand, and it has both sin 3 and ln(n + 2), this series n 2 # 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 : 1 1 1 sin ln(n + 2) · nsmall # = . 3 3 3 −( #) n 2 n 2 n 2 small

8 3 Because we can choose (small #) as small as we like and the power of n in the denominator is 2 −(small #), we expect this series to converge (we can choose (small #) so small that the exponent is > 1). In particular, 1 1 we can choose (small #) = 4 so that we will run limit comparison with bn = 5 . n 4 Let’s compute the relevant limit. One remark: I am going to be absurdly clever with how I group terms and compute the limit, but it really comes from the heuristics above. Also, anyway you want to compute the limit should be ﬁne!

1 1 sin 3 ln(n + 2) sin 3 ln(n + 2) lim n 2 = lim n 2 · . n→∞ 1 n→∞ 1 1 5 3 n 4 n 4 n 2 I will treat these two fractions separately. Note that

3 ln(n + 2) ln(x + 2) 4x 4 lim 1 = lim 1 = lim = 0 n→∞ n 4 x→∞ x 4 x→∞ x + 2 by L’Hopital’s rule, and likewise

1 1 sin 3 sin 3 1 lim n 2 = lim x 2 = lim cos = 1 n→∞ 1 x→∞ 1 x→∞ 3 3 3 x 2 n 2 x 2 by L’Hopital’s rule. Putting these together,

1 sin 3 ln(n + 2) n 2 lim 1 = 1 · 0 = 0. n→∞ 5 n 4

P∞ 1 Thus, we are in scenario (ii) of the limit comparison test. Because we know that the series n=1 5 con- n 4 P∞ 1 verges, we can conclude that the series n=1 sin 3 ln(n + 2) converges as well. n 2

Example 3.4. Determine whether the series

∞ X 1 √ e n n=1

converges or diverges.

Solution. There is no fluff to throw away, no logarithmic growth, and no way to use Taylor polynomials √ to our advantage (note that n → ∞ so it would not make sense to use a Taylor polynomial of ex, for example). This series will also deflect just about anything else you throw at it: it’s not geometric and cannot be converted into one, cannot be compared in any obvious way to a geometric series, and both the ratio and root tests are inconclusive. Because nothing else seems to work, this seems like a good Case IV infinite sum. My intuition tells me that the series should converge, just because it kind of looks geometric. Thus, let’s run limit comparison 1 against n2 and hope that it works out. Let’s compute:

√1 2 n n lim e = lim √ . n→∞ 1 n→∞ n n2 e

9 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 n2 = m4 and as n → ∞, m → ∞. The limit turns into

n2 m4 lim √ = lim . n→∞ e n m→∞ em Because we already know that ma bm for any a, b > 0, it follows that

m4 lim = 0. m→∞ em P∞ 1 Thus, we are in scenario (ii) of the limit comparison test and since we know that n=1 n2 converges, we ∞ P √1 can conclude that n=1 e n converges as well.

Example 3.5. Determine whether the series

∞ √1 X e n − 1 n n=1

converges or diverges.

√1 Solution. Because of the e n term, we can approximate the behavior in the numerator using Taylor polynomials. In particular, recall that the degree 1 Taylor polynomial centered at 0 for ex is 1 + x. Thus, we expect √1 x n √1 e − 1 ≈ x for x small and thus e − 1 ≈ n . Thus, we will run limit comparison against

√1 n 1 = 3 . n n 2 Note that √1 n √1 √1 e −1 n x e − 1 e − 1 √1 0 lim n = lim = lim = lim e x = e = 1 n→∞ √1 n→∞ √1 x→∞ √1 x→∞ n n x n by L’Hopital’s rule. Since we got a limit of 1, we are in scenario (i) of the limit comparison test. Since the √1 P∞ 1 P∞ e n −1 series n=1 3 converges, the series n=1 n converges as well. n 2

Example 3.6. Determine whether the series

∞ X 2n + ln n en + sin n n=1

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 2n, and 2n 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.

10 2n We will run limit comparison against en . Note that

2n+ln n ln n en+sin n 1 + 2n 1 + 0 lim n = lim = = 1. n→∞ 2 n→∞ sin n 1 + 0 en 1 + en The limit in the numerator comes from the fact that ln n na bn and the limit in the denominator is a sin n 1 result of 0 ≤ en ≤ en → 0. Because we got a limit of 1, we are in scenario (i) of the limit comparison test. Because ∞ ∞ n X 2n X 2 = en e n=1 n=1 P∞ 2n+ln n is a convergent geometric series (2 < e), we can conclude that the series n=1 en+sin n converges as well.

Example 3.7. Determine whether the series

∞ X ln(ln n) 1 5 4 n=1 (n + 1)

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) nsmall # 1 = . 1 5 5 −( #) (n5 + 1) 4 n 4 n 4 small

5 Because we can choose (small #) as small as we like, and the power of n in the denominator is 4 −(small #), we expect this series to converge (we can choose (small #) so small that the exponent remains larger than 1 1 1). In particular, we can pick small # = 8 and run limit comparison with 9 . Indeed, n 8

ln(ln n) 1 (n5+1) 4 ln(ln n) 1 1 lim 1 = lim 1 · 1 = 0 · 1 = 0. n→∞ n 8 n→∞ n 8 1 4 (1 + 0) 4 5 1 + n5 n 4 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

∞ 1 ∞ X n 8 X 1 5 = 9 4 8 n=1 n n=1 n

P∞ ln(ln n) converges, we can conclude that n=1 1 converges as well. (n5+1) 4