Chapter 8: Infinite Sequences and Series
Section 8.4: Power Series
Example 8.4.1
Determine the radius of convergence and the interval of convergence for the power series ∑n=1nn xn2nn+1!.
Even though (7) in Table 8.4.1 claims that absolute convergence at one end of the interval of convergence implies absolute convergence at the other, if the convergence at an endpoint is absolute, verify that it also absolute at the other.
Solution
Mathematical Solution
Since the given power series contains the powers xn, the radius of convergence is given by
R=limn→∞an/an+1 = limn→∞nnⅇ−nn+1!=2/e
At the right endpoint x=R, the given power series becomes Σ an Rn, which converges absolutely by part (1) of the Limit-Comparison test if the comparison series is taken as the convergent p-series Σ 1/n3/2.
The relevant calculation that must be made is c=limn→∞an Rn1/n3/2=1/2 π. Since c is a (finite) real number, both series converge. Hence, the power series converges absolutely at x=R.
At the left endpoint x=−R, the given power series becomes the alternating series Σ −1nan Rn, which has just been shown to converge absolutely.
Hence, the interval of convergence is −R,R=−2/e,2/e.
Maple Solution
Define the general coefficient an as a function of n
Write an=… Context Panel: Assign Function
an=nn2nn+1!→assign as functiona
Obtain the radius of convergence
Calculus palette: Limit template Context Panel: Assign Name
R=limn→∞anan+1→assign
Display R, the radius of convergence
Write R Context Panel: Evaluate and Display Inline
R = 2ⅇ−1
Test for convergence at x=R
Write An=… Context Panel: Assign Function
An=an⋅Rn→assign as functionA
Calculus palette: Limit template Context Panel: Evaluate and Display Inline
limn→∞An1/n3/2 = 122π
The comparison series Σ 1/n3/2 is a convergent p-series, so by part (1) of the Limit-Comparison test, the given series converges absolutely at x=R=2/e. The given series also converges absolutely at x=−R because an−Rn = An, and the series with those terms has already been shown to converge absolutely.
At the right endpoint x=R, the resulting series yields neither to the Ratio test (L=1) nor to the Integral test (Maple cannot evaluate the resulting integral). To find an appropriate comparison series, look for the limiting behavior of an Rn, which can be seen in the dominant term of its asymptotic series, obtained via the Context Panel as follows.
Write the expression. Context Panel: Evaluate and Display Inline
Context Panel: Series≻Asymptotic series≻n (In the Series dialog that opens, set "Series order" to 2.)
an Rn = nn2ⅇ−1n2nn+1!→asymptotic series in n122ⅇ−2ⅇ21n3/2π+O1n5/2
Maple's asympt command will also provide such an expansion, as shown in Table 8.4.1(a), below.
asymptan Rn, n,2 = 122ⅇ−2ⅇ21n3/2π+O1n5/2
Table 8.4.1(a) Asymptotic behavior of an Rn
The factor 1/n3/2 in the first term of the expansion is what suggests a comparison to the p-series Σ 1/n3/2.
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