<s>[INST] Power series expansion of $f(x)=\frac{x}{2x^2+1}$ I was requested to find a series representation of $$f(x)=\frac{x}{2x^2+1}$$
and its convergence radius. I attempted a solution but it is wrong according to online calculators, and I can't find my mistake.
My approach was the standard procedure of rewriting $f(x)$ as $\frac{1}{1-u}$ where $u$ is a function of $x$, and using the fact that $\frac{1}{1-u}=\sum_{n=0}^\infty u^n $. I noticed that
$$f(x)=\frac{x}{2x^2+1} = x \cdot \frac{1}{1-(-2x^2)} = x \sum_{n=0}^\infty (-2x^2)^n = x\sum_{n=0}^\infty (-2)^n (x^2)^n = \sum_{n=0}^\infty (-2)^n (x^2)^{n+1}$$
The expansion I get is therefore $f(x) = \frac{x}{2x^2+1}=x^2 - 2x^4 + 4x^6-8x^8+...$ with a convergence radius of $\frac{1}{\sqrt{2}}$.
As stated earlier, this does not match with the results I get from automated computations, and yet I can't find an error in my logic. I only started learning power series a couple of hours ago, so I must be clearly missing something. Any help would be much appreciated. Thanks in advance.
 [/INST] Another way to do it
$$f(x)=\frac{x}{2x^2+1}\quad \implies \int f(x)\,dx=\frac{1}{4} \log \left(1+2 x^2\right)$$ Let $t=2x^2$
$$\log(1+t)=\sum_{n=1}^\infty (-1)^{n+1} \frac {t^n}n$$
$$\frac{1}{4} \log \left(1+2 x^2\right)=\sum_{n=1}^\infty (-1)^{n+1} \frac {2^{n-2}}n x^{2n}$$
Now, differentiate
$$f(x)=\frac{x}{2x^2+1}=\sum_{n=1}^\infty (-1)^{n+1} 2^{n-1} x^{2 n-1}$$
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