Limit of a sequence given by recurrence relation and convergence rate

Question

Suppose we have a sequence $\{a_n\}_{n=0}^{\infty}$ which is generated by \begin{align*} a_{n+1} - \left(q+ \frac{A} {n+1} \right) a_n - \frac B n a_{n-1} = C, \end{align*} for $n \ge 1$, where $q, A, B, C$ are fixed positive constants and $0 < q < 1$.

Suppose the initial condition is $a_{0} = a_{1} = 1$. It is easy to see the sequence $\{a_n\}_{n=2}^{\infty}$ generated is positive and lower bounded by $C$. I want to argue the limit of the sequence exits and come up with the limit.

If we know there is a limit for the sequence, then by taking limits, it seems like $\lim_{n \to \infty} a_n = C/(1-q)$ and also this limit is not a fixed point of the recurrence. I guess the sequence will approximate this limit. But I cannot see how to argue the limit is guaranteed to exist. Any comments will be helpful. Thanks.

---

Update: I guess one way is to write the relation recursively, i.e., \begin{align*} a_{n+1} &= C + \left( q+\frac{A}{n+1} \right) a_n + \frac{B}{n} a_{n-1} \\ &= C + \left( q+\frac{A}{n+1} \right) \left( C + \left( q+\frac{A}{n} \right) a_{n-1} + \frac{B}{n-1} a_{n-2} \right) + \frac{B} {n} a_{n-1}, \end{align*} Since we want the limiting behavior, any term with $n$ in the denominator will vanish. Then if we keep expanding, we get \begin{align*} a_{n+1} = C + Cq + Cq^2 + \dots + C q^{n+1} + o(n). \end{align*} It seems right except it seems a bit too messy. If anyone has a neat proof, please point me out. Another question is whether we can estimate the convergence rate of $\{a_k\}$ to $C/(1-q)$. From the expansion the slowest mode seems to $AC/(n+1)$ and thus the rate is determined by this term. Is this correct? Thanks.

Answer 1

The sequence converges to $L=\frac C{1-q}$.

Proof: Put $a_n=L+b_n$. Then $b_n$, $n\geq0$, satisfy $$b_{n+1}=\left(q+\frac A{n+1}\right)b_n+\frac Bn b_{n-1}+\frac{AL}{n+1}+\frac{BL}n.$$

Claim 1: There exists a constant $K$ such that $|b_n|\leq \frac Kn$ for all $n\geq1$.

Proof of the Claim: We can choose $N\geq2$ so large that $q+\frac A{n+1}\leq Q$ and $\frac Bn\leq E$ for $n\geq N$, where $Q+E<1$. It can be chosen such that additionally $\frac{n+1}n\leq\frac{n+1}{n-1}\leq F$ for $n\geq N$, where $F$ is so close to 1 that we have $QF+EF<1$. We denote by $K_0$ the maximum of $n|b_n|$ for $n=1,\dots,N$.

Now we want to determine properties that $K$ must have such that the Claim can be proved by induction. If we choose $K\geq K_0$, there is no problem for $n\leq N$. Suppose now that the statement is true for $b_n$ and $b_{n-1}$ for some $n\geq N$. Then we estimate \begin{equation} \begin{matrix} (n+1)|b_{n+1}|&\leq &\left(q+\frac A{n+1}\right)(n+1)|b_n|+\frac Bn(n+1) |b_{n-1}|+{AL}+\frac{BL(n+1)}n\\&\leq&Q\,FK+E\,FK+AL+BLF.\end{matrix}\end{equation} IF $K$ satisfies $Q\,FK+E\,FK+AL+BLF\leq K$, then we can conclude that also $|b_{n+1}|\leq \frac K{n+1}$ and the induction is complete.

So the Claim holds if $K\geq K_0$ and $K\geq\frac{AL+BLF}{1-QF-EF}$. Here we need that $QF+EF<1$. Clearly such a $K$ can be chosen...

Edit:

We can prove more as the comments to the question suggest. Observe first that we can prove the following statement very similarly to the above Claim.

Claim 2: Consider a sequence $c_n$, $n\geq 0$, satisfying $$c_{n+1}=\left(q+\frac A{n+1}\right)c_n+\frac Bn c_{n-1}+r_n,\mbox{ where }|r_n|\leq R\, n^{-k}\mbox{ for }n\geq1$$ with a positive integer $k$ and some $R>0$. Then there exists a constant $K$ such that $|c_n|\leq K\,n^{-k}$ for $n\geq1$.

Now we prove by induction on $k$ that for all integers $k\geq 1$, there exists a polynomial $f_k(n)=L+C_1 n^{-1}+\cdots+C_{k-1}n^{-k+1}$ of $1/n$ and a constant $K$ such that $|a_n-f_k(n)|\leq K\,n^{-k}.$ Of course, the $C_k$ depend upon $A,B,C,q$, but we do not indicate this dependence.

For $k=1$, this is the contents of the Claim 1.

Suppose now that the statement is true for some integer $k\geq1$. For convenience, we introduce the operator $\newcommand{\LL}{{\mathcal L}}\LL$ by $$\LL(x)_n=x_{n+1}-\left(q+\frac A{n+1}\right)x_n-\frac Bn x_{n-1}\mbox{ for }n\geq1,\mbox{ if }x=(x_0,x_1,...).$$ We use the Landau symbol $O$: We write $O(n^{-g})$ to indicate any sequence $x_n$ for which there exists a constant $K$ such that $|x_n|\leq K n^{-g}$ for all $n\geq1$. Observe that $x_n=O(n^{-g})$ implies $\LL(x)_n=O(n^{-g})$.

By the induction hypothesis, we have $a_n=f_k(n)+O(n^{-k})$ for a certain positive integer $k$. Since $\LL(a)_n=C$ for all $n$, we have $\LL(f_k)_n=C+O(n^{-k})$. Now $f_k(n)$ is a polynomial in $1/n$ and therefore, by using Taylor expansions at $n=\infty$, $\LL(f_k)_n$ can be expanded in a convergent series in powers of $1/n$. By truncating this series, we find that $\LL(f_k)_n=C+h_k\,n^{-k}+O(n^{-k-1})$ with a certain coefficient $h_k$. Now we put $$f_{k+1}(n)=f_k(n)-\frac{h_k}{1-q}n^{-k}$$ and find that $\LL(f_{k+1})_n=C+O(n^{-k-1})$. This implies that $\LL(a-f_{k+1})_n=O(n^{-k-1})$. Now we can apply Claim 2 and obtain that $a_n-f_{k+1}(n)=O(n^{-k-1})$ as wanted. This completes the induction.