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Sum of series/ ln(1+2x+3x^2)

Sum of series ln(1+2x+3x^2)


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n = 1
$$\sum_{n=1}^{\infty} \log{\left(3 x^{2} + \left(2 x + 1\right) \right)}$$ 
Sum(log(1 + 2*x + 3*x^2), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\log{\left(3 x^{2} + \left(2 x + 1\right) \right)}$$
It is a series of species
$$a_{n} \left(c x - x_{0}\right)^{d n}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}$$
In this case
$$a_{n} = \log{\left(3 x^{2} + 2 x + 1 \right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} 1$$
Let's take the limit
we find
True

False
The answer [src] 
      /             2\
oo*log\1 + 2*x + 3*x /
$$\infty \log{\left(3 x^{2} + 2 x + 1 \right)}$$ 
oo*log(1 + 2*x + 3*x^2)

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