Kenneth Kuttler

250 CHAPTER 11. POWER SERIES

Furthermore, the above discussion shows this series solves the initial value problem on itsinterval of convergence. It only remains to show the radius of convergence of this seriesequals 1. It will then follow that this series equals (1+ x)α because of uniqueness of theinitial value problem. To find the radius of convergence, use the ratio test. Thus the ratioof the absolute values of (n+1)st term to the absolute value of the nth term is∣∣∣α(α−1)···(α−n+1)(α−n)

(n+1)n!

∣∣∣ |x|n+1∣∣∣α(α−1)···(α−n+1)n!

∣∣∣ |x|n = |x| |α −n|n+1

→ |x|

showing that the radius of convergence is 1 since the series converges if |x|< 1 and divergesif |x|> 1.

The expression, α(α−1)···(α−n+1)n! is often denoted as

(α

). With this notation, the follow-

ing theorem has been established.

Theorem 11.4.2 Let α be a real number and let |x|< 1. Then

(1+ x)α =∞

∑n=0

(α

)xn.

There is a very interesting issue related to the above theorem which illustrates the limi-tation of power series. The function f (x) = (1+ x)α makes sense for all x >−1 but one isonly able to describe it with a power series on the interval (−1,1) . Think about this. Theabove technique is a standard one for obtaining solutions of differential equations and thisexample illustrates a deficiency in the method.

To completely understand power series, it is necessary to take a course in complexanalysis. It turns out that the right way to consider Taylor series is through the use ofgeometric series and something called the Cauchy integral formula of complex analysis.However, these are topics for another course.

11.5 Exercises1. Verify the power series claimed in the chapter for cos(x) ,sin(x) and ex. The method

for doing this was shown in the chapter in the case of ex. Go through the detailscarefully and then do the same details for cos(x) ,sin(x).

2. The logarithm test states the following. Suppose ak ̸= 0 for large k and that p =

limk→∞

ln(

1|ak|

)lnk exists. If p > 1, then ∑

∞k=1 ak converges absolutely. If p < 1, then

the series, ∑∞k=1 ak does not converge absolutely. Prove this theorem.

3. Using the Cauchy condensation test, determine the convergence of ∑∞k=2

1k lnk . Now

determine the convergence of ∑∞k=2

1k(lnk)1.001 .

4. Find the values of p for which the following series converges and the values of p forwhich it diverges.

∞

∑k=4

1lnp (ln(k)) ln(k)k

250 CHAPTER 11. POWER SERIESFurthermore, the above discussion shows this series solves the initial value problem on itsinterval of convergence. It only remains to show the radius of convergence of this seriesequals 1. It will then follow that this series equals (1 +.x)~ because of uniqueness of theinitial value problem. To find the radius of convergence, use the ratio test. Thus the ratioof the absolute values of (n +1)” term to the absolute value of the n‘” term isa(@—1)---(@—n+1)(a—n)(n+1)n!a(a—1)--(a—n+1) | |x|”|x|"*!|or — nln+1= |x| + |aln!}showing that the radius of convergence is | since the series converges if |x| < 1 and divergesif |x| > 1.The expression, alan t)fa~nt)ing theorem has been established.is often denoted as (“) . With this notation, the follow-Theorem 11.4.2 Let @ be a real number and let |x| < 1. Then(4y%=¥ (“yxn=o \7tThere is a very interesting issue related to the above theorem which illustrates the limi-tation of power series. The function f (x) = (1 -+x)* makes sense for all x > —1 but one isonly able to describe it with a power series on the interval (—1,1). Think about this. Theabove technique is a standard one for obtaining solutions of differential equations and thisexample illustrates a deficiency in the method.To completely understand power series, it is necessary to take a course in complexanalysis. It turns out that the right way to consider Taylor series is through the use ofgeometric series and something called the Cauchy integral formula of complex analysis.However, these are topics for another course.11.5 Exercises1. Verify the power series claimed in the chapter for cos (x) , sin (x) and e*. The methodfor doing this was shown in the chapter in the case of e*. Go through the detailscarefully and then do the same details for cos (x) , sin (x).2. The logarithm test states the following. Suppose a, ¢ 0 for large k and that p =nar)limy_50.—,z exists. If p > 1, then Ye, a, converges absolutely. If p < 1, thenthe series, 7°, ax does not converge absolutely. Prove this theorem.3. Using the Cauchy condensation test, determine the convergence of ))"_5 cur: Now. oo 1determine the convergence of )°_5 inky rOOT4. Find the values of p for which the following series converges and the values of p forwhich it diverges.y 1£, In? (In (k)) In (k)