What is the radius of convergence of $a_n$ where $a_{n+1}=frac{n-5}{n+1}a_n$?

Question

Consider a sequence ${a_n}$ which satisfies
$$a_{n+1}=frac{n-5}{n+1}a_n$$

What is the radius of convergence of $sum _n a_n x^n$?

Clearly, $a_n=0$ for all $ngeq 6$.

André Armatowski · Accepted Answer

Assuming you start at $n=0$ with $a_{0}=kinmathbb{R}$, your sum  is the polynomial
$$k-5kx+10kx^{2}-10kx^{3}+5kx^{4}-kx^{5}$$
which "converges" for all $xin mathbb{R}$ (Convergence barely makes sense here, since it is clearly finite for each $xinmathbb{R}$). Therefore the radius of convergence is $infty$.

Oliver Diaz · Answer

That is correct, and so the series is actually a polynomial. Thus the radius of convergence is $infty$, that is the "series" converges for $|x|<infty$  (all $x$)
Although this is an overkill, one can also use the formula for the radius of convergence:
$$ R=frac{1}{limsup_nsqrt[n]{|a_n|}}=frac{1}{0}=infty$$
Not surprisingly we get the same.

What is the radius of convergence of $a_n$ where $a_{n+1}=frac{n-5}{n+1}a_n$?

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