Radius of convergence of $\sum_{n=0}^\infty a_n x^n$, with $a_{n+2} = \frac{n(n+1) a_{n+1} - a_n}{(n+2)(n+1)}, a_2 = -a_0/2$

Question

Problem

Find the radius of convergence of the power series

$$ \sum_{n=0}^\infty a_n x^n $$

where $a_n$'s are defined by the following recurrence relation

$$ \begin{aligned} a_{n+2} &= \frac{n(n+1) a_{n+1} - a_n}{(n+2)(n+1)}, n\ge 1 \\[8pt] a_2 &= -a_0/2 \end{aligned} $$

with arbitrary $a_0, a_1$.

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Try1

I have tried to directly applying the ratio test, by dividing the above recurrence relation by $a_{n+1}$,

$$ r_{n+1} = \frac{n}{n+2} - \frac{1}{(n+2)(n+1)} \frac{1}{r_n} $$

where $r_n := a_{n+1}/a_n$. We can observe that

$$ r_n r_{n+1} = \frac{n}{n+2} r_n - \frac{1}{(n+2)(n+1)} $$

but I cannot proceed to find the expression for

$$ \lim_{n \to \infty} \vert r_n \vert $$

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Try2

Since the recurrence relation depends on the arbitrary choice pf $a_0, a_1$, let us proceed by letting $a_0:= 0$. We have

$$ \begin{aligned} a_2 &= a_0 = 0 \\[7pt] a_3 &= -\frac{1}{6} a_1 \\[7pt] a_4 &= \frac{2}{4} a_3 = -a_1/12 \\[7pt] a_5 &= \frac{3}{5} a_4 - \frac{1}{5\cdot 4}a_3 = -\frac{1}{24} a_1 \\[7pt] \end{aligned} $$

where I cannot find any simple rules. Likewise, by letting $a_1:= 0$, we have

$$ \begin{aligned} a_2 &= -a_0/2 \\[7pt] a_3 &= \frac{1}{3} a_2 = -\frac{1}{6}a_0 \\[7pt] a_4 &= \frac{2}{4} a_3 - \frac{1}{4\cdot 3}a_2 = -\frac{1}{24} a_0\\[7pt] a_5 &= \frac{3}{5} a_4 - \frac{1}{5\cdot 4}a_3 = -\frac{1}{60} a_0 \\[7pt] \end{aligned} $$

where again I have failed to find any rules. So, I cannot find the interval that the following composition is valid.

$$ \sum_{n=0}^\infty a_n x^n = a_0 \left[ 1 - x^2/2 - x^3/6 - x^4/24 - x^5/60 + \cdots \right] + a_1 \left[ x - x^3/6 - x^4/12 - x^5/24 + \cdots\right] $$

Any help will be appreciated.

Answer 1

The trick is to notice $a_{n+2}=-\frac{f_n}{(n+2)!}(a_0+a_1)$ for $n\ge1$ with $f_1=f_2=1,\,f_n=nf_{n-1}-f_{n-2}$. Asymptotically $f_n\sim nf_{n-1}$, so $\frac{f_n}{(n+2)!}$ is a sequence in which the ratio of consecutive terms $\to1$. Thus the radius of convergence is also $1$.

Answer 2

A possible way to approach the problem is to consider $$F(x):=\sum_{n\ge0}a_nx^n$$ as a purely formal series (so without thinking about convergence radii etc) and try to understand what function does it actually represent. Playing around with the recursive formula for the coefficients, you can derive (up to mistakes on my part) $$\frac{d^2}{dx^2}F(x) = -F(x)+x\frac{d^2}{dx^2}F(x)\ .$$ From this, you can try to solve the differential equation to find a closed formula for $F(x)$, derive the Taylor coefficients from it, and thus determine the $a_n$. Having done this, you can finally determine the convergence radius.