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The question is: Determine the interval of convergence of the power series

My attempt at an answer: $$u_n=\frac{(2n+1)}{(n^2+1)}(2x+1)^{12}$$ Applying the ratio test: $$\begin{align} \require{enclose} \frac{|u_{n+1}|}{|u_n|}&=\left|\frac{(2n+2)(2x+1)^{12}}{((n+1)^2+1)}.\frac{(n^2+1)}{(2n+1)(2x+1)^{12}}\right|\\ &=\left|\frac{(2n+2)\enclose{horizontalstrike}{(2x+1)^{12}}}{(n^2+2n+1)}.\frac{(n^2+1)}{(2n+1)\enclose{horizontalstrike}{(2x+1)^{12}}}\right|\\ &=\left|\frac{(2n+2)}{(n^2+2n+1)}.\frac{(n^2+1)}{(2n+1)}\right|\\ \end{align}$$ But now I just got rid of all the $x$ components which is obviously wrong!?!.

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This is not a power series. Typo? – André Nicolas Nov 5 '12 at 20:27
up vote 8 down vote accepted

If the series really is $$\sum_{n=1}^\infty\left(\frac{2n+1}{n^2+1}\right)(2x+1)^{12}\;,$$ the loss of $x$ is obviously right: this is simply


which is either $a(2x+1)^{12}$ for $$a=\sum_{n=1}^\infty\frac{2n+1}{n^2+1}$$ if that series converges, or undefined if it does not. And in fact it’s undefined, since

$$\sum_{n=1}^\infty\frac{2n+1}{n^2+1}$$ is readily seen to diverge, e.g., by limit comparison with the harmonic series.

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Except when $\: x = -\frac12 \:$. $\;\;$ – Ricky Demer Nov 5 '12 at 20:47
@Brian M. Scott: can you explain the following in a bit more detail? which is either $a(2x+1)^{12}$ for $a=\sum_{n=1}^\infty\frac{2n+1}{n^2+1}$ Is that $a$ times $(2x+1)^{12}$? If it's on that side of equals shouldn't it be under the line? – Gineer Nov 6 '12 at 6:32
@Gineer: Yes, it is, and no, it should not. The point is that if it existed, $\sum_{n=1}^\infty\frac{2n+1}{n^2+1}$ would by a constant; I simply called that constant $a$, so that $\left(\sum_{n=1}^\infty\frac{2n+1}{n^2+1}\right)(2x+1)^{12}=a(2x+1)^{12}$. – Brian M. Scott Nov 6 '12 at 13:01

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