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I was trying to calculate the limit of sequence defined as

$$a_1=k; a_2=l; a_{n+1}=\frac{a_n+(2n-1)a_{n-1}}{2n}; k, l\in\mathbb{N}k<l$$

I had no idea where to start with that so I've brute forced the problem on my PC for permutations of $(k, l)$ to $a_{10^{10}}$ in hope that there would be an emerging pattern. This is what my PC thinks the $a_\infty$ is for $(k, l)$:

$$(1, 2) \Rightarrow 3-\sqrt{2}$$ $$(1, 3) \Rightarrow 5-\sqrt{8}$$ $$(2, 3) \Rightarrow 4-\sqrt{2}$$ $$(1, 4) \Rightarrow 7-\sqrt{18}$$ $$(2, 4) \Rightarrow 6-\sqrt{8}$$ $$(3, 4) \Rightarrow 5-\sqrt{2}$$ $$....$$

It seems that $a_{n\rightarrow\infty}\rightarrow2l-k-(l-k)\sqrt{2}$. How can I show that mathematically, without using brute force?

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  • $\begingroup$ I doubt it. If $a_n$ converges to a single value, when setting $a_{n+1}=a_n=a_{n-1}$ you get $a_{n+1}=1$. $\endgroup$ – Yves Daoust Dec 21 '15 at 13:27
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    $\begingroup$ @YvesDaoust For any $a_{n+1}=a_n=a_{n-1}$, $a_n=\frac{a_n+(2n-1)a_n}{2n}=\frac{2na_n}{2n}=a_n$ It can converge to any number, depending on first two members of the series $\endgroup$ – Mirac7 Dec 21 '15 at 13:30
  • $\begingroup$ Ooooops, right, sorry. $\endgroup$ – Yves Daoust Dec 21 '15 at 13:31
  • $\begingroup$ It's a sequence, not a series. $\endgroup$ – Hans Lundmark Dec 21 '15 at 13:46
  • $\begingroup$ Your conjectured formula doesn't match your particular values; it should be $(l-k)\sqrt2$ rather than $(l-k+1)\sqrt2$. $\endgroup$ – Henning Makholm Dec 21 '15 at 16:41
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We have $$a_{n+1} - a_n = \dfrac{a_n+(2n-1)a_{n-1}}{2n} - a_n = \dfrac{(2n-1)a_{n-1}-(2n-1)a_{n}}{2n} = -\dfrac{2n-1}{2n} \left(a_n-a_{n-1}\right)$$ Let $b_n = a_{n+1}-a_{n}$. We then have $$b_{n+1} = - \dfrac{2n-1}{2n}b_n$$ with $b_1 = l-k$. We then have $$b_{n+1}=b_1(-1)^{n} \prod_{k=1}^n \dfrac{2k-1}{2k} = b_1(-1)^n \dfrac{(2n)!}{4^n(n!)^2} = b_1\left(-\dfrac14\right)^n \dbinom{2n}n$$ We have $$a_{n+1} -a_1= \sum_{k=1}^n \left(a_{k+1}-a_k\right) = \sum_{k=1}^nb_k = b_1 \sum_{k=0}^{n-1} \left(-\dfrac14\right)^k \dbinom{2k}k$$ Recall that $$\sum_{k=0}^{\infty} \dbinom{2k}k x^k = \dfrac{1}{\sqrt{1-4x}}$$for $\vert x \vert \leq 1/4$. Plugging in $x=-1/4$, we obtain $$\lim_{n \to \infty} a_{n+1} = a_1 + \dfrac{b_1}{\sqrt2}=k+ \dfrac{l-k}{\sqrt2}$$

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  • $\begingroup$ Your solution doesn't seem to correspond with what I got. However, I don't see an error in your solution, but my solution has been calculated by a computer with significant degree of precision and therefore shouldn't have errors with such deviation from your solution. $\endgroup$ – Mirac7 Dec 21 '15 at 13:56
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    $\begingroup$ @Mirac7 I had a index mix up. Even then, it doesn't converge to your answer. Note that if $l=k$, then we need the limit to be $k$, which is not the case with your answer. So I doubt if your answer is true. $\endgroup$ – Leg Dec 21 '15 at 15:29
  • $\begingroup$ The formula had an error which I fixed, and while it's true that it's just my conjecture, and doesn't work for $k=l$, the limits above should be correct as they were calculated, one member of the sequence at a time, and they converge to $\langle1.58578,1.58579\rangle$ for $(k, l)=(1,2)$, and your result $1+\frac{1}{\sqrt{2}}$ does not; similar for other values. I am completely confused, as I double checked your solution and it seems correct to me, but I also triple checked that my calculations correctly approximate to very high precision. $\endgroup$ – Mirac7 Dec 22 '15 at 9:24
  • $\begingroup$ It appears to me that you made a mistake at $b_{n+1}=\frac{2n-1}{2n}b_n$. I believe it should be $b_n=\frac{2n-1}{2n}b_{n-1}$. $\endgroup$ – Mirac7 Dec 28 '15 at 11:24

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