My answer comes out to be $[-1,1]$. Also,if the question asks for interval of convergence, do we have to write the interval of convergence or the interval of absolute convergence? Cause, as per my knowledge both are different, and it should be specified in the question.
The series $\sum_{n=1}^{\infty}\frac{x^{n-1}}{n^2}$ converges absolutely for $x∈(-1,1)$?
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real-analysis
power-series
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0i think in this case, the intervals are the same? – 2017-01-14
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0the radius of convergence of a power series is the radius of the largest (open) disk in which the series converges. – 2017-01-14
1 Answers
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$$\sum_{n=1}^{\infty}\frac{x^{n-1}}{n^2}$$
- The ratio/root test tell you that the radius of convergence of this series is $1$
- So, for $|x|>1$, the terms do not go to $0$, and the series will diverge there
- For $|x|\le 1$, note that $\left| \frac{x^{n-1}}{n^2}\right| \le\frac{1}{n^2}$, which has a convergent sum. Hence, in this region, the sum is absolutely convergent, and hence convergent in the regular sense also.
We thus conclude that this series converges (absolutely, and in the standard sense) for $|x|\le1$.