Chapter 8: Infinite Sequences and Series
Section 8.4: Power Series
Example 8.4.14
Determine the radius of convergence and the interval of convergence for the power series ∑n=2∞xnn lnn.
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→∞1n lnn1n+1 lnn+1=limn→∞n+1nlnn+1lnn=1
At the right endpoint x=R=1, the given power series becomes Σ an=∑n=21n lnn, which diverges by the Integral test since fx=1x lnx decreases monotonically to zero on 2,∞ and its integral on that interval diverges.
At the left endpoint x=−R=−1, the given power series becomes the alternating series Σ −1nan, which converges conditionally by the Leibniz test since the sequence an decreases monotonically to zero.
Hence, the interval of convergence is −R,R=−1,1.
Maple Solution
In Figure 8.4.14(a), the graph of
fx=1x lnx
is drawn in red, while the graph of its derivative
f′x=−1+lnxx2lnx2
is drawn in green. Since 1+lnx>0 for x>1/e, it sould be clear that fx decreases monotonically to zero as x→∞.
module() local f,p; f:=1/x/ln(x); p:=plot([f,diff(f,x)],x=2..10,color=[red,green],labels=[x,y],tickmarks=[8,3]); print(p); end module:
Figure 8.4.14(a) Graph of f (red) and f′ (green)
Define the general coefficient an as a function of n
Write an=… Context Panel: Assign Function
an=1n lnn→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 = 1
Test for convergence at x=R=1
Calculus palette: Definite integral template Context Panel: Evaluate and Display Inline
∫2∞1x lnx ⅆx = ∞
Test for convergence at x=R=−1
Write an⋅−Rn Context Panel: Evaluate and Display Inline
an⋅−Rn = −1nn⁢ln⁡n
At the right endpoint x=R=1, the given power series becomes Σ an=∑n=21n lnn, which diverges by the Integral test. Figure 8.4.14(a) and the related discussion reveal that an=fn, and fx is an appropriate function to use as an integrand. Since the integral diverges, so too does the series.
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