Convergence tests These notes discuss a number of tests for determining whether a series converges or diverges.

0.1. Geometric Series Formula. This is in your book, but I thought it might be helpful to include here. If you have a geometric series

1 ark = a + ar + ar2 + ar3 + ... Xk=0 with a = 0 then this converges if and only if r < 1. If we do have r < 1thenthegeometric 6 a | | | | series converges to 1 r .

n 1 Problem 0. Does the series k=1 3k converge? Justify your answer and compute its value. Solution: n 1 = 1 + 1 + 1 + ... is a geometric series with a = 1 and r = 1 .Since k=1 3k 3 32 P 33 3 3 r < 1, we see that the series converges to a = 1/3 = 1/3 = 1 by the Geometric Series | | P 1 r 1 1/3 2/3 2 Formula.

0.2. Integral test. Consider the sum k1=1 ak.Letf(x)beafunctionsuchthat: f(x)ispositiveanddecreasingforallP x>0and • f(k)=a for all k =1, 2,...,. • k Then:

If 1 f(x)dx is finite then 1 a converges. • 1 n=1 n If 1 f(x)dx is not finite then 1 a diverges. • R1 P n=1 n R 1 P 1 Problem 1. Show that k=1 k diverges. Solution: Let f(x)= 1 . Note that f(x)satisfiesthehypothesesoftheIntegralTest,namely: x P f(x)ispositiveanddecreasingforallx>0, and • f(k)= 1 for all k =1, 2, 3,.... • k 1 1 We may thus apply the Integral Test. We recall from Chapter II that 1 x dx is not finite. (If you had forgotten this, it comes from either “the chart” for improper integrals, or from adirectcomputation.) R 1 1 1 1 Since 1 x dx is not finite, the Integral Test shows that the series k=1 k diverges. R P 1 1 Problem 2. Show that k=1 k2 converges. 1 Solution: Let f(x)= x2 . Note that f(x)satisfiesthehypothesesoftheIntegralTest, namely: P f(x)ispositiveanddecreasingforallx>0, and • f(k)= 1 for all k =1, 2, 3,.... • k2 1 1 We may thus apply the Integral Test. We recall from Chapter II that 1 x2 dx is finite. If you had forgotten this, it comes from either “the chart” for improper integrals, or from the 1 R direct computation: b 1 1 1 dx =lim dx x2 b x2 Z1 !1 Z1 1 b =lim x b 1 !1 ⇥ 1 ⇤ =lim +1 =1. b b !1  1 1 1 1 Since 1 x2 dx is finite, the Integral Test shows that the series k=1 k2 converges.

Non-exampleR of integral test! To use the integral test, it is essentialP that you verify all of the hypotheses. Watch what happens if you work with a function f(x)bethatisnot positive and decreasing. We consider k1=1 sin(⇡ k)andf(x) = sin(⇡ x). For the integral we have: · · P b 1 1 sin(⇡x)dx =lim cos(⇡ x) b ⇡ · Z1 !1  1 which does not exist. By contrast:

1 sin(⇡ k) = sin(⇡)+sin(2⇡)+sin(3⇡)+ =0+0+0+ =0. · ··· ··· Xk=1 The Integral Test fails in this case because f(x)isnot a positive and decreasing function. Moral: always check the hypotheses of any test you are using!

1 1 1 0.3. The kp test. The series k=1 kp converges if and only if p>1. This isn’t really a separate test. It is just an immediate consequence of the integral test, 1 1 P and the fact that 1 xp dx converges if and only if p>1. R 0.4. Term test. Consider the series k1=1 ak. If limn an =0then 1 ak diverges. • !1 6 k=1 P However: if limn an =0,thenthetestisinconclusiveandyoumustusesomeother • test. !1 P

k2 1 Problem 3. Show that k=1 3k2+k+1 diverges. Solution: We will use the Term Test. We compute P n2 lim an =lim n n 2 !1 !1 3n + n +1 1 2 2 n =lim n n 1 3n2 + n +1 !1 ✓ n2 ◆ 1 1 1 =lim = = n 1 1 !1 3+ n + n2 3+0+0 3

1 k2 Since the limit of the terms is =0,theTermTestshowsthattheseries 1 diverges. 3 6 k=1 3k2+k+1 P 0.5. Ratio test. Consider a series k1=1 ak.Computethelimitingratio:

P a L := lim n+1 n a !1 n If L<1then an converges. • If L>1then an diverges. • If L =1orifthelimitdoesnotexist,thenthetestisinconclusiveandyoumustuseP • some other test.P

n n3 Problem 4. Show that ( 1) 3n converges. Solution: We will use the Ratio Test. Compute P a (n +1)3/3n+1 1 (n +1)3 1 L := lim n+1 =lim =lim = . n a n n3/3n n 3 n3 3 !1 n !1 !1 1 Since L = 3 < 1, the Ratio Test shows that the sum converges.

Problem 6. Show that the Taylor series for ex converges for any fixed value x = b.1 Solution: If we take T 1ex and plug in x = b we get:

1 xk 1 bk T 1ex x=b = x=b = | k! ! | k! Xk=0 Xk=0 We want to use the Ratio Test. So we compute the limit:

bn+1/(n +1)! bn+1 n! b L =lim =lim =lim =0. n n n n n !1 b /n! !1 b · (n +1)! !1 n +1

Since L =0< 1, the Ratio Test implies that this series converges.

nn Problem 7. Show that n! diverges. Solution: We will use the Ratio Test. For this, we compute the limiting ratio: P (n +1)n+1/(n +1)! (n +1)n+1 (n +1)n 1 n L =lim =lim =lim =lim 1+ = e. n nn/n! n (n +1)nn n nn n n !1 !1 !1 !1 ✓ ◆ Since L = e>1theRatioTestshowsthattheseriesdiverges.

1 Note that showing that the Taylor series T 1ex x=b converges is not quite the same as showing that the sum converges and that the limit is eb. We will discuss| this in Chapter V.5. 1 1 Non-example of Ratio Test. Consider the series k=1 k2+5k .Ifwetrytheratiotest,we get: 1 P 2 L =lim (n+1) +5(n+1) n 1 !1 n2+5n n2 +5n =lim n 2 !1 (n +1) +5(n +1) n2 +5n =lim n 2 !1 n +2n +1+5n +5 n2 +5n =lim n 2 !1 n +7n +6 2 1 Since the biggest term in sight is n we scale top and bottom by n2 to get: 1/n2 n2 +5n =lim n 1/n2 n2 +7n +6 !1 ✓ ◆ 1+ 5 1+0 =lim n = =1 n 7 6 !1 1+ n + n2 1+0+0

Since L =1,theratiotestisinconclusive. 1 1 Thus, to determine whether the series k=1 k2+5k converges or diverges, we would need to use another test. (Hint: the Limit Comparison Test or the Integral Test.) P n 0.6. Root test. Consider the sum k1=1 ak. Define L := an . (Note that there is an absolute value under the square root!) | | P p If L<1then a converges. • n If L>1then an diverges. • If L =1orifthelimitdoesnotexist,thenthetestisinconclusiveandyoumustuseP • some other test.P Note: The root test is most useful when the nth term already has an nth power in it.

n 1 1 Problem 9. Show that n=1 sin ( n )converges. n Solution: Recall that sinn( 1 )= sin( 1 ) . Since there is already an nth power in the nth P n n term, we will use the Root Test. For this, we compute L =lim n sinn( 1 ) =lim n sin( 1 ) n =lim sin( 1 ) = sin(0) =0. n | n | n | n | n | n | | | !1 q !1 q !1 Since L =0< 1theRootTestshowsthatthisseriesconverges.

n n3 1 Problem 10. Show that n=1 3n3+5n2+7 converges. Solution: We want to use the Root Test. For this, we must compute: P ⇣ ⌘ n3 n n3 1 L =lim n =lim = . n 3n3 +5n2 +7 n 3n3 +5n2 +7 3 !1 s✓ ◆ !1 1 Since L = 3 < 1, the Root Test implies that this series converges. 0.7. Limit Comparison Test. This is very similar to the test we saw for improper integrals.

Let k1=0 ak and k1=0 bk be two series. Assume that ak > 0andbk > 0forallk,and P• anP limn exists and equals a positive number. (In particular, this limit cannot equal • !1 bn 0or !) 1 Then either both series converge or both series diverge.

n2 1 Problem 11. Determine where n=1 an converges, where an = n3+4n+11 . Justify your answer. Solution: We will use the LimitP Comparison Test and the Integral Test. As n ,the largest terms in the numerator and denominator will dominate. So we compare!1 this to n2 1 1 n=1 bn where bn n3 = n .Wenotethat: an > 0foralln,sinceboththenumeratoranddenominatorallpositiveforalln,and P • 1 bn > 0foralln since n is positive for all n. Also: • a n2 n lim n =lim n b n n3 +4n +11 · 1 !1 n !1 ✓ ◆ n3 =lim n 3 !1 n +4n +11 =1 So the limit exists and equals a positive number. Thus we may apply the Limit Comparison Test to conclude that either both series converge or both series divert. 1 1 1 Next we use the Integral Test to analyze k=1 k .Letf(x)= x .Wenote: f(x)ispostivieanddecreasingforallx>0andthat P • f(k)= 1 for all k =1, 2, 3,.... • k 1 1 1 Thus we may apply the Integral Test. Since 1 f(x)dx = 1 x dx does not exist, the Integral 1 Test implies that 1 diverges. By the Limit Comparison Test, this in turn implies that: k=1 k R R n2 1 the original series,P n=1 n3+4n+11 also diverges.

n3+pn+7 P 1 Problem 12. Show that n=1 an converges when an = 4n5+11n3/2+14 . n3 1 Hint: first use the Limit Comparison Test with bn = 4n5 = 4n2 .ThenusetheIntegralTest 1 P 1 to see that n=1 4n2 converges. Problem to discuss: what test would you use? P sin(1/k)k • k2+11 k+3k2 • P k2+11 • P ek+k 1 . • Pk2+5k