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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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    Except when $\: x = -\frac12 \:$. $\;\;$2012-11-05
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    @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?2012-11-06
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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