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I am asked to find the power series of the function $f(x)=\arctan(\frac{x}{\sqrt{2}})$. I first found the derivative of this function which is: $f'(x)=\frac{\sqrt{2}}{2+x^{2}}$. Then I found the power series of $f'(x)$ which is: $\sum_{n=0}^{\infty }\frac{1}{\sqrt{2}}(-1)^{n}\left ( \frac{1}{2} \right )^{n}x^{2n}$. Then to get the power series of $f(x)$, I just integrated the previous power series term by term to get: $\sum_{n=0}^{\infty }\frac{1}{\sqrt{2}}(-1)^{n}\left ( \frac{1}{2} \right )^{n}\frac{x^{2n+1}}{2n+1}$. Now to find the radius of convergence, I found $-\sqrt{2}\leqslant x\leqslant \sqrt{2}$. However, when I entered this answer in the software, it doesn't accept this answer as a valid answer. Can anybody tell me where the problem is? Because I checked and I found that the series is convergent at the end points of my interval of convergence. Thanks!

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The radius of convergence of a power series is a non-negative number, that can have the value of $\infty$. You should try putting $R = \sqrt{2}$ into the software. Note that you should integrate power serieses only within the radius of convergence. Your answer is still correct event though strictly speaking not correctly established.

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  • $\begingroup$ But in some cases the radius of convergence is $\infty$. For example: $\displaystyle\sum_{n=0}^\infty\frac{z^n}{n!}$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 12 '13 at 15:55
  • $\begingroup$ True, I corrected it. The edit wasn't mine. It was originally in the form "a non-negative number accepting the value of $\infty$. $\endgroup$ – Juha-Matti Vihtanen Jun 13 '13 at 13:24
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The ratio test tells you the series converges if $-\sqrt{2}<x<\sqrt{2}$ and diverges if $x<-\sqrt{2}$ or $x>\sqrt{2}$.

But what happens when $x=\text{exactly }\sqrt{2}$ or $-\sqrt{2}$?

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  • $\begingroup$ Is he abel to figure this out? $\endgroup$ – ncmathsadist Jun 12 '13 at 1:51
  • $\begingroup$ You should dirichlet the OP do it on his/her own. $\endgroup$ – Pedro Tamaroff Jun 12 '13 at 1:54
  • $\begingroup$ @Michael Hardy: I checked both cases, when $x=-\sqrt{2}$ and when $x=\sqrt{2}$ and I got an alternating series which is convergent by the alternating series test, but still the software is saying that my answer is wrong. $\endgroup$ – Alex K Jun 12 '13 at 1:58
  • $\begingroup$ The alternating series test works when the terms decrease in absolute value and their limit is $0$ and they alternate. In this case, their limit is not $0$, so that doesn't work. When $x=\sqrt{2}$, you get a series of positive terms all equal to each other so that sum diverges to $\infty$. And when $x=-\sqrt{2}$, they alternate in sign but they are all equal in absolute value, so the series diverges but doesn't go to $\infty$ or $-\infty$ or anything like that. $\endgroup$ – Michael Hardy Jun 12 '13 at 2:05

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