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The question is: Determine the interval of convergence of the power series
$\sum_{n=1}^\infty\left(\frac{2n+1}{n^2+1}\right)(2x+1)^{12}$

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!?!.
$_{help!!!}$

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    This is not a power series. Typo?2012-11-05

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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

$(2x+1)^{12}\sum_{n=1}^\infty\frac{2n+1}{n^2+1}\;,$

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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    @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 $\left(\sum_{n=1}^\infty\frac{2n+1}{n^2+1}\right)(2x+1)^{12}=a(2x+1)^{12}$.2012-11-06