MA 146 Steve Sawin

11.4 - Limit Comparison and Comparison Test

The Limit Comparison Test will be our workhorse test. It relies on an idea we have explored before, which is the idea of the “large x behavior” of a function or equivalently the “large n behavior” of a sequence. If you do not feel comfortable with that, I will be posting a recorded lecture called “Large x behavior” to review and explain it. The simple version of this test says this. When k is very large 1/k2 and 1/(k2 + 1) are P∞ 1 P∞ 1 practically the same. So since k=1 k2 CONVERGES, it must be the case that k=1 k2+1 converges. To say this with enough care that you can be confident of the answer requires some work though. ak Limit Comparison Test: If ak and bk are two positive sequences and limk→∞ 6= 0 bk but exists, then if ∞ X bk k=0 CONVERGES then ∞ X ak k=0 CONVERGES and if the first DIVERGES the second DIVERGES. Steps to use LCT:

1. Find the simplified sequence of terms bk – Write ak, throw away all lower order terms in top and bottom and simplify until it looks like a p-series, Geometric Series or some other series you know. P 2. Find the convergence/divergence of the simplified series bk – Use the p-Series Test or the Geometric Series Test to decide what the simplified sum does.

3. Confirm that the ratio approaches 1 (or any number but 0) – Write out the ratio, and take the limit carefully to make sure your reasoning was correct. This will always work out easily as long as your thinking in (1) was solid P 4. Conclude that k ak CONVERGES or DIVERGES – This is the payoff. First Example: Does the following series converge or diverge.

∞ X k + 1 • k2 + ln(k) k=1 Answer: 1. Throw away lower order stuff k + 1 k + 0 1 ∼ ∼ k2 + ln(k) k2 + 0 k

because 1 is lower order than k and ln(k) is lower order than k2. We end with a p-series, so we are good.

2. The simplified series P 1/k is a p-series with p = 1 (if this is not easy for you yet, practice it a bit until it is). This means by the p-Series Test this simplified series diverges.

3. Eat your vegetables: we have to check the LCT can be used. Put original terms on top, simplified on bottom.

(k + 1)/(k2 + ln(k)) k(k + 1) lim = lim k→∞ 1/k k→∞ k2 + ln(k) k2 + k k2 = lim = lim = 1 k→∞ k2 + ln(k) k→∞ k2 Notice we had a ratio of fractions and had to do that “flip over the denominator” trick. Since this limit is 1, this worked.

4. By the LCT, the original series P(k + 1)/(k2 + ln(k)) does the same thing as the simplified series, which by (2) means it DIVERGES.

Second Example: Does the following series converge or diverge.

∞ X k2 + 2k • k − 5(3)k k=1 Answer:

1. Throw away lower order stuff

k2 + 2k 0 + 2k 2k 1 2k ∼ = = − k − 5(3)k 0 − 5(3)k −5(3)k 5 3

because k2 is lower order than 2k and k is lower order than 3k. Notice we did a little algebra (including laws of exponents) to write the final expression as something we recognize as Geometric. As soon as you saw k in an exponent somewhere you knew that is where we would end up!

P 1 2
k 2. The simplified series − 5 3 is a Geometric Series with r = 2/3 (keep practicing! You’ll get it!). This means by the GST (Geometric Series Test) this simplified series converges. 3. Check! Put original terms on top, simplified on bottom.

(k2 + 2k)/(k − 5(3)k) (−5)3k(k2 + 2k) lim = lim k→∞ (−1/5)(2/3)k k→∞ 2k(k − 5(3)k) (−5)(k23k + 6k) −5(6)k = lim = lim = 1 k→∞ k2k − 5(6)k) k→∞ −5(6)k

I used a lot of laws of exponents in there, Since this limit is 1, this worked.

4. By the LCT, the original series P(k2 + 2k)/(k − 5(3)k) does the same thing as the simplified series, which by (2) means it CONVERGES.

Examples: Use LCT to decide if these series converge or diverge

∞ √ X n + 1 • n2 + 1 n=1

Answer: Simplify: √ √ n + 1 n n1/2 1 ∼ = = n2 + 1 n2 n2 n3/2 which is a p-series with p = 3/2. The simplified series converges by the PST (p-Series Test) so check √ √ √ √ ( n + 1)/(n2 + 1) n3/2 n + 1 n4 + n3/2 n4 lim = lim = lim = lim = 1 n→∞ 1/n3/2 n→∞ n2 + 1 n→∞ n2 + 1 n→∞ n2

So LCT says the original series CONVERGES.

∞ X k3/2 + 2k4 + ln(k) • k + 2k5 k=1 ∞ ∞ X n − n2 X 5n − n2 • • 4n5 + 3n2 2n + 3 n=1 n=0 ∞ √ X 4k + 3k2 • 2k + (1/2)k k=1

The Comparison Test is to the Limit Comparison Test as driving stick shift is to auto- matic: In the hands of an expert it can do everything the other plus a little more and a little better, but it takes a long time practicing before you are not stalling out in traffic. Comparison Test - The Statement: Suppose ak and bk are positive terms and for P∞ P∞ P∞ each k ak ≤ bk. If k=0 bk CONVERGES then k=0 ak CONVERGES, and if k=0 ak P∞ DIVERGES then k=0 bk DIVERGES. This doesn’t tell you how to use it. P∞ Comparison Test - The Tool: Suppose your series is k=0 ak and all the ak are P∞ positive. If you can dream up a series k=0 bk that is bigger (bk ≥ ak) and converges, then P∞ your series CONVERGES. If you can dream up a series k=0 bk that is smaller (bk ≤ ak) and diverges, then your series DIVERGES. First Example: Does the series ∞ X ln(k) • k k=1 converge or diverge? Answer: Notice LCT does not help because you cannot simplify the top and bottom, and DT (Divergence Test) does not help because the sequence of terms ak converge to 0. You look at this and ln(k) is growing pretty slow so you figure it will not matter as much as 1/k. If it were 1/k this would be p-series with p = 1 so it would diverge. It is not, but that is OK, because it is BIGGER than the p-series which diverges. Specifically ln(k) > 1 as long as k > e, so ln(k)/k > 1/k when k is at least 3. Remember it is good enough of the condition holds eventually, i.e. after the first few terms of the series. Since ln(k)/k is greater than 1/k which diverges, P ln(k)/k DIVERGES. Second Example: Does the series ∞ X 1 • k! k=1 converge or diverge? Answer: Again, the individual terms go to zero so the DT tells us nothing. There is nothing to simplify so LCT tells us nothing. But we know k! grows faster than any exponential, so the terms are getting small very fast, so it seems like the sum should converge. Since it grows faster than any exponential we compare it to, say 2k. We have 0! = 20, 1! < 21, 2! < 22, 3! < 23, but 4! > 24 and after that 5! is 5 times as big as 4! while 25 is only twice as big as 25, so the left hand side is going to keep being bigger than the right as k grows. So we see k! > 2k for k ≥ 4. That means 1 1 < k! 2k P k for k at least 4. This is helpful because k 1/2 is Geometric with r = 1/2 so it converges by the GST. Since P 1/k! is less than a convergent series, it is CONVERGENT. Examples: Use comparison series to determine if the following series converge or diverge ∞ X 1 X • Compare to 1/k ln(k) k=2 ∞ X sin2(k) X • Compare to 1/k2. k2 k=1