This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 12. 4 The Root and Ratio Tests '..‘ Fe}; 1:. The root and ratio tests are amongst the most powerful tests. They are both based on comparing a given series to a geometric series
Ens". Recall that the geometric series converges for iml < 1 and diverges for :1:[ 2 1. Observe too that
lim (risky/k
[ﬁt—700 =15. Theorem 12.4.1 (Page 593 of SHE) The Root Test
Let 0.1g 2 0 for all k. Assume that lim al/k=p. k—roo (a) If p < 1, Zak converges. (b) If p > 1, Eek diverges. (If p = 1, the test is inconclusive). Proof (a) We assume ,0 < 1. Choose 11. such that p<p<1. For sufﬁciently large k, we have milk<p => ck < 111“.
Since (0 <);1 < I, E 15’“ is a convergent geometric series. By the basic comparison test, 2 a]: converges.
(b) is similar. I Remark
We often use the root test when our terms a]; are of the form something raised to a kth power. Example 1
Test for convergence SO k 1/}:
lim cal/k lim i
k—boo k k—mo 111k , 1
[claim—“1 By the root test, k
EEK; = 2 (milk) converges. Example 2
Test for convergence
kmo
3k
Solution
Here kmo
Gk — W
19100 1/»’c
1 I:
kmo/k
: 3
(kl/k)100
= ——3——.
Recall that
lim 331/9“ = 1 => klim 191/,“ = 1.
m—J'OO #00
So
ﬁm lfk _ 1' (k1/k)100 _ 1 < 1
Jar+00 Gk H4 kings 3 ~ 3 ' By the root test, the series
kIOO ZR}; 2 Z @ converges. Example 3
Test for convergence 216“. 59 Example 3
Test for convergence Zkk. Solution
Here Ogle/k =k—roo,k——roo.
That is [)2 lirn ailk=oo>L law—’00 By the root test, the series diverges. The root test works well with powers. The ratio test works well with facto
rials: Theorem 12.4.2 (Page 595 of SHE) The Ratio Test
Let (1;; > 0 for all 16. Suppose that lim ““1 = A.
J‘s—«+00 G};
(a) If A < 1, then Eek converges.
(b) If A > 1, then 205;. diverges.
(If A = 1, the test is inconclusive).
Proof
One uses the basic comparison test, showing, very roughly1 that 04; is compara—
ble to A’“. I Example
Test for convergence
4k
"is?
Solution
Here
4]:
[Bk 3
so
4k+1 4k
akil/ak = [(kl—lﬂlH]
k!
2 4
(k + 1)'
1 .
= 4—.
k + 1
Then 4
saw = .122. m = 0 <1 60 By the ratio test, the series converges. Example
Test for convergence (3k)!
2 100K Solution
Here
_ (3k)!
“k _ look
so
(3 (k + 1))! (3k)!
“HI/“’4 : mow/W
_ (31: + 3)! 100’“
_ (3k)! 100Ic+1
1
= k _
(3 + 3) (3k + 2)(3k + 1) 100
—> 00,16: —> 00.
That is, lim ak+1/ak = 00 > 1. k—roo By the ratio test, the series diverges. SUMMARY OF TESTS SO FAR
We have been dealing with series 2 a}: with positive terms.
(I) Usually, when a series has terms with powers, e.g. 2k k2
21+?6 or 227 but no factorials, we use the root test.
(II) When there are also factorials, e.g. 4!:
Zn, we use the ratio test.
(III) When the terms have the form of a numerator and denominator with powers of k, e.g.,
Z W2 + 2
4k? 4 x/E ’
we use the limit comparison test (or basic comparison test) and compare to a
p—series Z [cl—P. 61 12.5 Absolute, Conditional Con
vergence So far we have dealt mainly with series with positive terms. Now we study
series Zak where different terms may have different signs. Very often, we can
reduce this to studying Z Iakl . Deﬁnition Absolute Convergence
If E jak converges, then we say that
Zak converges absolutely. The reason is that this is useful is: Theorem 12.5.1 (page 597 of SHE) Absolute convergence implies con vergence
If Zions] converges,
then Zak converges.
Proof
Now —le S are S lakl
#OSak+lak S2akl. By the comprison test, since 2 ak converges, also 2 (2 ak) = 2 E ak eonw
verges, so Z: (a;c + lak) converges.
Then Zak = Z[(ak+lakl)—»lakll
= ZWHGkDZlakl is a difference of two convergent series, so Eek converges. I Example
Show that if p > 1, , k
i (— 1) converges.
kl"
[0:1 62 Solution Here k
(—1)  1
Gk: kp‘:]aki:g;'
As p > 1, we know that
DO 00 1
Z Iakl = E E converges.
Ic=1 k=1
That is,
°° °° (—1)‘°
Z (1;, = Z: converges absolutely, and so converges.
k?
k=1 lc=1
Remark We shall soon see that for any 19 > 0
00 I:
Z( 1) converges, even though it does not converge absolutely for p S 1. We shall call this condi—
tional convergence: Deﬁnition Conditional Convergence
Suppose that 2 oh converges,
but
Zlak} diverges. Then we say that 2 a1, converges conditionally. Theorem 12.5.3 (Page 598 of SHE) Alternating Series Test
Let me > 0 for k 2 0. Assume that
(i) {ck} is a decreasing sequence, i.e. no 2 al 2 a2 2 ...; (ii) lim ilk = 0.
k—rroo
. Then Do
2 (—4)!“ (1k converges.
k=0
Remark We call series of this form alternating series.
Idea of Proof 63 The series converges because succesive terms cancel one another, due to opposite
sign. There is a beautifui proof in SHE: Let m 2 1, and consider 32m, the even
order partial sum. We see that 2m 2 (—1)ka k=0
= (a0 — £11) + (a2  a3) + + (a2mv2 m a2m—1)+ 62m 52m Here each term in brackets, namely
00 — G1; 82 — as; ; Gem—2 r a2me1
is positive (recaﬂ r11 < a0 and (13 < (12 etc). So
32m>0form21. On the other hand, _1)2m+1 32m+2 E 32m + ( ﬂ2m+1 + a2m+2 = 82m ' a2m+1 + a2m+2 < 32m: since a2m+1 > a2m+2. Thus {5%,} is a decreasing sequence of positive numbers.
As it is bounded below (by 0), we have lim 82m = L Til—>00 for some L > 0. Then also the odd order partial sums converge to the same
limit: 2m+1
S2m+1 = 52m+('—1) a2m+1 52m _“ aZm+1 Hence
lim 32m+1 = lim (82m ” 32m+1)
773‘qu m—‘OO : L — 0 = L.
Hence hm sm = L. m—roo I
Example Let p > O. Showr that °° Hf“
2 hp converges.
k=1 64 For which 39 > 0 does it converge absolutely? For which p > 0 does it converge conditionally?
Solution
llere 00 _1k
£8; has the form 00
2 (Uk We:
16:1
Where 1
a5, 2 E is positive, decreasing (as k increases)
and I
13520 Gk E 0‘ By the alternating series test, °° (—lr °°
2 k1” = Z (—1)"c a}, converges.
[6:1 k=1 We already saw that for p > 1, the series converges absolutely. If p 5 1, then 00 00 1
Z ‘(—l)kekl : 2 E) diverges,
k=1 Ic=1 as it is a p—series, with p S 1. Thus 00 (_ 1)}:
2 k1” converges conditionally for p g l.
k=1 Summary When you see a series with terms having alternating sign, you can try apply the
alternating series test, provided you can easily check that the terms decrease
in magnitude, and have limit 0. In many cases, however, it is easier to take
absolute values, and test for absolute convergence. If there is absolute conver— gence, then there is convergence. 65 ...
View
Full Document
 Spring '08
 Morley

Click to edit the document details