The limit of a recurrance relation

Question

Consider the linear recurrence relation given by

$a_{k+2} = a_{k+1} + \frac{1}{k}a_k$

and note that $a_k=k$ solves this recurrence. Consider the limit $L$ defined as

$L = \displaystyle\lim_{k\to\infty}\frac{1}{k}a_k$.

As noted, with the initial condition $a_1=1, a_2=2$ we have that $a_k=k$ and thus the limit $L$ equates to $1$. For different initial conditions however, it is not clear to me how to calculate this limit. In particular, I'd like to know what the limit is for $a_1=a_2=1$.

More generally, I'd like to know where I should look to read more about how to calculate recurrences with polynomial coefficients.

FYI, using an Approach0 search, there's the related 2 conjectured recursion limits for $e$ and $\pi$., which is closed as a duplicate of Mirror algorithm for computing $\pi$ and $e$ - does it hint on some connection between them?. There's also the AoPS thread convergence 2. — John Omielan, Aug 26 '24 at 05:53
@JohnOmielan, Thanks a lot. These reference are perfect. In particular this one: https://math.stackexchange.com/questions/1664933/mirror-algorithm-for-computing-pi-and-e-does-it-hint-on-some-connection-b.
Please let me know if you'd like me to close this question, or allow you to write this as an answer. If you'd like to close it, feel free to do so yourself.

Thanks again and have a good day.

Mathew — Mathew, Aug 26 '24 at 06:08
You're welcome. I'm glad those references helped you. In terms of you closing this question, that's up to you. Regarding me writing an answer based on your linked question, thanks for the offer. However, I think it would instead perhaps be better that you answer your own question, such as by including that question link and highlighting any particular parts of it, and any of its answers, that you found most useful. — John Omielan, Aug 26 '24 at 15:28

score 1 · Accepted Answer · answered Aug 26 '24 at 07:28

Too long for a comment.

Considering $$a_{k+2} = a_{k+1} + \frac{1}{k}a_k\qquad \text{with} \quad a_1=\alpha \quad \text{and}\quad a_2=\beta$$ Using the methods and results from the different linked posts given in comments $$a_k=\alpha k+\frac{(\beta -2 \alpha )\, \Gamma(k+1,-1)}{e\, \Gamma (k)}$$

Using asymptotics to get more than the limit itself $$\frac {a_k} k=\frac{(e-2) \alpha +\beta }{e}+(-1)^k\, \frac{e^k (\beta -2 \alpha )}{k^{k+\frac{3}{2}}\sqrt{2 \pi }}+O\left(\frac{e^k}{k^{k+\frac{5}{2}} } \right)$$

Trying for $\alpha=\beta=1$ and $k=10$ $$\frac {a_{10}}{10}=\frac{28319}{44800}$$ while, converted to decimals, the above gives $\color{red} {0.63212053}10$

The limit of a recurrance relation

1 Answers1