Limit of a sequenced defined by arithmetic mean and geometric mean.

Question

If $a_1 =\alpha$, $ a_2 = \beta$, and, for every $k$, $$a_{2k+1}= \frac{1} {2k} \sum_{i=1}^{2k} a_i\qquad a_{2k+2}=\left(\prod_{i=1}^{2k+1} a_i\right)^{1/(2k+1)}$$ what is the limit of the sequence $(a_k)$?, That is, what is $\lim_{n\to\infty}{a_n}?$

I think this sequence converges, but i can't find a clue to figure out its limit.

When $\alpha=\beta$, it is clear that $a_n=\alpha$ However, if $\alpha\neq\beta,$ I think i cannot calculate exactly wht $a_n$is, so I'm currently stuck.

I know that for all $k$, $\alpha \le a_k \le \beta$, and I think that this sequence converges towards somewhere close to $\min(\alpha, \beta)$ because $a_1=\alpha$, $a_2=\beta$, $a_3=\frac{\alpha + \beta}{2}$, $a_4=(\frac{\alpha\beta(\alpha+\beta)}{2})^{1/3}$, but I can't figure out what to do next.

For example, here is a sequence when $\alpha=2, \beta=4$: $$a_1=2, a_2=4, a_3=3, a_4=2.8845, a_5=2.9711, a_6=2.90162, a_7=2.95953, a_8=2.9098, a_9=2.95331, a_{10}=2.91462, a_{11}=2.9494, a_{12}=2.91776, a_{13}=2.9468, a_{14}=2.91998...$$ Edit: I think as $n$ increases, $$a_{2k+2}\le a_{2k+4}\le a_{2k+3}\le a_{2k+1}$$ Could anybody confirm this please? If this is right, $a_n$ converges to $2.9.....$

I found that the above inequality is right using mathematical induction

Let $\alpha<\beta$, then $a_3=\frac{\alpha+\beta}{2}$, $a_4=\sqrt[3]{\alpha\beta\frac{\alpha+\beta}{2}}=\sqrt[3]{(\frac{\alpha+\beta}{2}+\frac{\beta-\alpha}{2})(\frac{\alpha+\beta}{2}-\frac{\beta-\alpha}{2})(\frac{\alpha+\beta}{2})}=\sqrt[3]{(\frac{\alpha+\beta}{2})^3-A}<\frac{\alpha+\beta}{2}=a_3$ If $a_{2k+1}>a_{2k+2}$, $$a_{2k+3}=\frac{1}{2k+2}\sum_{i=1}^{2k+2}{a_i}=\frac{a_{2k+2}+a_{2k+1}+\sum_{i=1}^{2k}{a_i}}{2k+2}=\frac{a_{2k+2}+(2k+1)a_{2k+1}}{2k+2}\\\therefore a_{2k+2}<a_{2k+3}<a_{2k+1}\\a_{2k+4}=(\prod_{i=1}^{2k+3}{a_i})^{\frac{1}{2k+3}}=(a_{2k+3}\cdot (a_{2k+2})^{2k+2})^{\frac{1}{2k+3}}\\\therefore a_{2k+2}<a_{2k+4}<a_{2k+3}\\\therefore a_{2k+2}<a_{2k+4}<a_{2k+3}<a_{2k+1}$$By the mathematical induction, for all natural number $n$, $a_{2n+2}<a_{2n+4}<a_{2n+3}<a_{2n+1}$

This means that the sequence $\{a_{2n}\}$ and $\{a_{2n+1}\}$ converges, since the two sequences are monotone increasing and monotone decreasing, and the two sequences are bounded(since $\alpha<\{a_{2n}\}<\{a_{2n+1}\}<\beta$). Now I want to know whether the two limits of these two sequences are the same, which means $\{a_{n}\}$ converges.

Answer 1

Both $a_{2n}$ and $a_{2n+1}$ seem to approach a limit $L$ with a speed $L+C/n$. One way to speed convergence is to eliminate $C$ and $n$ from three consecutive terms: $$ x = L+\frac C{n-1}\\y=L+\frac Cn\\z=L+\frac C{n+1}\\ x-2y+z=\frac{2C}{n^3-n}\\ 2xz-yx-yz=\frac{2CL}{n^3-n}\\ L\approx \frac{2xz-yx-yz}{x-2y+z}$$ In the following diagram, the + and the x graphs have been fitted from the $a_{2n}$ sequence and the $a_{2n+1}$ sequence.

The limit seems to 2.93288, before underflow causes problems.

EDIT:

If $a_1=1-X$ and $a_2=1+X$, then $a_3=1$ and all the rest of the $a_n$ are even functions of $X$. They all approach $1$ as $X\to0$, so I tried $$a_n=1+b(n)X^2+c(n)X^4+d(n)X^6+\cdots (n\geq4)$$ By ignoring higher-order terms, it seems that $$b(4)=-1/3\\b(5)=-1/3+1/4\\b(6)=-1/3+1/4-1/5\\b(7)=-1/3+1/4-1/5+1/6$$ and this pattern continues. In the limit as $n\to\infty$, $b(n)\to\frac12-\ln2$. So an approximation to the limit, when $a_1=1-X$ and $a_2=1+X$, is $$1-(\ln2-\frac12)X^2$$ For example, if $X=\frac13$ then $a_1=\frac23$ and $a_2=\frac43$, and the limit is $1-(\ln2-\frac12)/9$. Then, when you triple $a_1$ and $a_2$, so you start with $2$ and $4$, the approximation is $$3-(\ln2-\frac12)/3\approx 2.9356$$ which is out by $0.003$.

Answer 2

I don't offer any analytical insight, but I hope some numerical experiments would be useful.

One thing I have to warn about: this sequence converges very slowly, in fact, so slowly, that without unlimited precision software (like Mathematica) it's very hard to find even $3$ digits of the limit. I might try this in Mathematica next week, but so far I have used R for experiments.

(To the OP: R is free, you can download and install it, and copy the following script there if you want to do some experiments yourself).

To make the recurrence easier to program, I have reformulated it the following way:

$$\begin{cases} a_0=a \\ b_0=b \\ \displaystyle s_0=\frac{a+b}{2} \\ p_0 = (ab)^{1/3} \end{cases}$$

$$\begin{cases} a_n=s_{n-1} \\ \displaystyle b_n=p_{n-1} a^{\frac{1}{2n+1}}\\ \displaystyle s_n=\frac{n}{n+1}s_{n-1}+\frac{a_n+b_n}{2n+2} \\ \displaystyle p_n = p_{n-1}^{\frac{2n+1}{2n+3}} \left( a_n b_n \right)^{\frac{1}{2n+3}} \end{cases}$$

There might be better ways to do it, but this one came to mind. All the numbers seem to converge to the same limit (I have no formal proof for it though).

For $n=100$ and for the initial data the OP used ($a=2,b=4$) the results are:

$$a_n=2.9337 \dots \\ b_n=2.9320 \dots$$

We can assume that the limit is about $2.933 \dots$, but it's hard to say more without using unlimited precision.

The convergence is easy to see numerically, though I agree with the OP that the fact that $\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n$ remains to be proved.

Here's the plot for $a_n$ (red), $b_n$ (blue), $s_n$ (brown) and $p_n$ (green) in the above case, generated by R:

Here's the program in R:

#Kind of AGM, but different
options(digits=22)
a <-2;
b <- 4;
a0 <- a;
b0 <- b;
s0 <- (a0+b0)/2;
p0 <- (a0*b0)^(1/3);
n <- 0;
N <- 100;
x <- 1:N;
y <- rep.int(a,N);
z <- rep.int(b,N);
u <- rep.int(s0,N);
w <- rep.int(p0,N);
while (n < N-1){
                n <- n+1;
                a1 <- s0;
                b1 <- p0*a1^(1/(2*n+1));
                s1 <- n*s0/(n+1)+(a1+b1)/(2*n+2);
                p1 <- p0^((2*n+1)/(2*n+3))*(a1*b1)^(1/(2*n+3));
                y[n+1] <- a1;
                z[n+1] <- b1;
                u[n+1] <- s1;
                w[n+1] <- p1;
                a0 <- a1;
                b0 <- b1;
                s0 <- s1;
                p0 <- p1;
}
y2 <- (a+b)/2;
y1 <- (a*b*(a+b)/2)^(1/3);
plot(x,y,col="red",ylim=c(y1,y2),xlab="n",ylab="an,bn");
points(x,z,col="blue")
points(x,u,col="brown")
points(x,w,col="green")
a1
b1

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Concerning the convergence proof. Taking the limit for the recurrence relations I used we have:

For $n \gg 1$:

$$\begin{cases} a_n=s_{n-1} \\ b_n=p_{n-1} \\ s_n=s_{n-1} \\ p_n = p_{n-1} \end{cases}$$

Because:

$$\lim_{ n \to \infty} x^{1/n}=1$$

$$\lim_{ n \to \infty} \frac{n}{n+1}=1$$

$$\lim_{ n \to \infty} \frac{x}{n}=0$$

This may not be completely rigorous, but it's clear enough.