Math · Calculus II · Worked example
Find the interval of convergence
Find the interval of convergence of the power series of xⁿ/n, starting at n = 1.
Apply the ratio test
The ratio of consecutive terms is n|x|/(n + 1), which approaches |x|. So the series converges when |x| < 1 and diverges when |x| > 1: the radius is 1.
Test x = 1
The series becomes the harmonic series, which diverges.
Test x = −1
The series becomes −1 + 1/2 − 1/3 + ⋯, an alternating series with terms shrinking to 0, so it converges.
State the interval
Closed at −1 and open at 1.
Result
[−1, 1).
Your turn
Find the interval of convergence of the power series of xⁿ/n!, starting at n = 0.
Show the answer and explanation
All real numbers.
The ratio of consecutive terms is |x|/(n + 1), which approaches 0 for every x, so the series converges everywhere. Its sum is eˣ.
Keep exploring
In Sequences & infinite series, the terms at x = 1 form the p-series with p = 1: after twenty terms the partial sum is 3.60 and still growing, and the studio classifies it as divergent.
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