<s>[INST] Power Series: Ratio test, but $x$ cancels out?!? 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!?!.
$_{help!!!}$
 [/INST] 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.
</s>