I want to compute the terms $a_n$ from some recursion formula of the form:
$a_n=\sum_{k=1}^nb_ka_{n-k}$ with some initial condition $a_0$, and the $b_k$ are known. Also, let's assume there is no closed form for $a_n$.
These terms grow very rapidly. Therefore I thought using a recursion formula involving the logs might be more tractable. However, I don't know if it is possible to go from this recursion formula involving $a_n$ and $b_n$ to another one involving $\log(a_n)$ and $\log(b_n)$ only. At first it seemed obvious to me, but I am somewhat doubting now.
The main problem I encounter there is that I get the log of a sum on the right-hand side, that I would like to express as a sum of the logs instead. Maybe some kind of good approximation is possible ?
Some context : this problem occurs in a physics problem, when computing recursively some partition function. I want to avoid overflow issues.
EDIT : adding some information:
If it is of any help, $b_k=(\frac{1}{1-e^{-k\beta}})^3$ where $\beta$ can be close to $0$ (so $b_k$ can possibly take very high values). The initial condition is $a_0=1$.
If $a_n =\sum_{k=1}^nb_ka_{n-k} $, let $A(x) =\sum_{n=0}^{\infty} a_n x^n $ and $B(x) =\sum_{n=0}^{\infty} b_n x^n $.
Then, in the usual manner,
$\begin{array}\\ A(x)B(x) &=\sum_{i=0}^{\infty} a_i x^i\sum_{j=0}^{\infty} b_j x^j\\ &=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty} a_i b_j x^{i+j}\\ &=\sum_{n=0}^{\infty}\sum_{i=0}^{n} a_i b_{n-i} x^{n}\\ &=\sum_{n=0}^{\infty}x^n\sum_{i=0}^{n} a_{n-i} b_{i} \\ &=\sum_{n=0}^{\infty}x^n\sum_{i=0}^{n} a_{n-i} b_{i} \\ &=a_0b_0+\sum_{n=1}^{\infty}x^n\left(a_nb_0+\sum_{i=1}^{n} a_{n-i} b_{i}\right)\\ &=a_0b_0+\sum_{n=1}^{\infty}x^na_nb_0+\sum_{n=1}^{\infty}x^n\sum_{i=1}^{n} a_{n-i} b_{i}\\ &=a_0b_0+b_0\sum_{n=1}^{\infty}x^na_n+\sum_{n=1}^{\infty}x^n\sum_{i=1}^{n} a_{n-i} b_{i}\\ &=a_0b_0+b_0(A(x)-a_0)+\sum_{n=1}^{\infty}x^n a_{n}\\ &=a_0b_0+b_0(A(x)-a_0)+(A(x)-a_0)\\ &=A(x)-a_0\\ \text{so}\\ a_0 &=A(x)-A(x)B(x)\\ &=A(x)(1-B(x))\\ \text{or}\\ A(x) &=\dfrac{1}{1-B(x)}\\ \end{array} $
If you can write $1-B(x) =\prod_{k=1}^m (c_k-x) $, you can get an estimate for the growth of the $a_n$ in terms of the largest of the $(1/c_k)^n$.
If $B(x)$ is a power series rather than a polynomial, it is harder.