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I'm going through old exams for my Calc III course and came across a problem that I did not know how to do. The problem is:

Find the interval of convergence of the series

$\sum_{n=0}^{\infty}\frac{(x-2)^{n^2}}{2n+1}$

and determine whether the series converges absolutely or conditionally at the endpoints.

I can't think of any tests that would work. The Ratio test doesn't yield anything solvable and the Root Test seems also useless here. The Comparison Test might work but I'm having trouble finding a series to compare it to.

Thank you in advance for any help you can give.

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I like this problem! I will have to assign this sometime.

Apply the root test. The denominator satisfies $\lim_{n \to \infty} \sqrt[n]{2n+1} = 1$ (use L'Hopitals). The limit in the numerator simplifies to $\lim_{n \to \infty} |x-2|^{n} ,$ which is 0 when $|x-2|<1.$ Therefore the preliminary interval of converge is $1<x<3.$ The series is divergent at the right endpoint $x=3$ (p-series,p=1) and convergent at $x=1$ (Alt. Series Convergence Test can be used here)

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  • $\begingroup$ I'm glad you enjoyed it. Quick question: I don't follow how the right end point is divergent ($x=3$). If $x$ is $3$ then the numerator is always 1, and the denominator goes to infinity, so shouldn't it converge to 0? $\endgroup$ – Kommander Kitten Apr 23 '15 at 18:33
  • $\begingroup$ No, for $\sum a_n$ to converge, we must have $\lim a_n=0.$ However, the converse is not true. For your example, $x=3$ leads to $\sum 1/(2n+1).$ Use the integral test: $\int_0^\infty \frac{1}{2x+1} \ dx = 0.5 \ln(2x+1) |^{x \to \infty}_0 = \infty - 0 $ $\endgroup$ – matt biesecker Apr 23 '15 at 18:57

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