Section 10.14 : Power Series
5. For the following power series determine the interval and radius of convergence.
∞∑n=16nn(4x−1)n−1Show All Steps Hide All Steps
Start SolutionOkay, let’s start off with the Ratio Test to get our hands on L.
L=limn→∞|6n+1(4x−1)nn+1n6n(4x−1)n−1|=limn→∞|6n(4x−1)n+1|=|4x−1|limn→∞6nn+1=6|4x−1| Show Step 2So, we know that the series will converge if,
6|4x−1|<1→24|x−14|<1→|x−14|<124 Show Step 3So, from the previous step we see that the radius of convergence is R=124.
Show Step 4Now, let’s start working on the interval of convergence. Let’s break up the inequality we got in Step 2.
−124<x−14<124→524<x<724 Show Step 5To finalize the interval of convergence we need to check the end points of the inequality from Step 4.
x=524:∞∑n=16nn(−16)n−1=∞∑n=16nn(−1)n−16n−1=∞∑n=16(−1)n−1nx=724:∞∑n=16nn(16)n−1=∞∑n=16nn16n−1=∞∑n=16n
Now, the first series is an alternating harmonic series which we know converges (or you could just do a quick Alternating Series Test to verify this) and the second series diverges by the p-series test.
Show Step 6The interval of convergence is below and for summary purposes the radius of convergence is also shown.
Interval:524≤x<724R=124