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Today I saw the following problem,

Let $a_1=1$ and $a_{n+1}=a_n+\frac1{a_n^2}$. Find the integer part of $a_{900}$.

The following solution was given,

We can use induction to prove that for $n\ge2$ we have $$\sqrt[3]{3n+2}\le a_n\le\sqrt[3]{3n+\frac{31}{15}+\ln\frac{3n-1}5-\frac1{9n-3}}.$$ Plugging in $n=900$ we obtain $13.9282<a_{900}<13.9391$.

Clearly this solution has extremely high precision. However, I am confused about how it is obtained. I suppose the left side (lower bound) is clear: we have $$a_{n+1}^3=a_n^3+3+\frac3{a_n^3}+\frac1{a_n^6}\ge a_n^3+3.$$ Considering $a_2=2=\sqrt[3]8$, we get the form $\sqrt[3]{3n+2}$.

In comparison, the expression on the right is horrendous. My question is: in general, is there a technique we can use to estimate a recursive sequence using a function containing roots, fractions, polynomials, logarithms, exponents? You may use the derivation of the upper bound in this problem to illustrate the method, if it exists.

Michael Rozenberg
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youthdoo
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  • I think that a similar issue appeared in solving this problem: https://math.stackexchange.com/questions/4881759/prove-sum-limits-i-12024a-i314-where-a-1-2-a-i-2-sin-fraca-i-12 But here we need a more powerful technique/tool to deal with a variety of recursive relations. – youthdoo Jun 14 '25 at 07:58
  • See general algorithm in my answer here. My Mathematica code with input $f(x)=x+\frac{1}{x^2}$ produced output $n\sim G(a_n)$, where $G(x)=\frac{1}{3}x^3-\log x+\alpha+O(x^{-3})$, inverse this asymptotic gives $a_n\sim\sqrt[3]{3n+\log n}$ – Quý Nhân Jun 14 '25 at 08:29
  • I couldn't really understand how your algorithm worked, but how to determine $\alpha$, and why does your result show $\mathrm O\left(x^{-3}\right)$? @QuýNhân – youthdoo Jun 14 '25 at 09:00
  • $\alpha$ depends on seed $a_1$ and there is no closed form for it. See this answer by Will Jagy for full explanation, I coded based on his instruction. Oh may be you asked why $O(x^{-3})$ and not $O(x^{-1})$? As I mentioned we can't find $\alpha$ exactly, thus the expression in your post only evaluate correctly up to before constant term (in the root) – Quý Nhân Jun 14 '25 at 09:29
  • @QuýNhân I will try to understand that post. By the way, your program does not return anything if $f(x)=2\sin\frac x2$. – youthdoo Jun 14 '25 at 10:38
  • @yoothdoo No, the requirement is $f(x)\sim x+O(1)$ (Laurent series not Taylor series), and $a_n$ diverges. Your input in this case must be $\frac{1}{2\sin(\frac{1}{2x})}$. Btw, there is no efficient algorithm to find full expansion of functional inversion, so you have to calculate asymptotic of $a_n$ manually – Quý Nhân Jun 14 '25 at 10:51
  • The trick using lower and upper bounds is widely used. FYI, I hope to see solutions using this trick for my question here and here – River Li Jun 14 '25 at 14:22
  • @RiverLi It's a bit different again. The recursive relation has got "$n$" as well as $a_n$. – youthdoo Jun 14 '25 at 16:45
  • @youthdoo I think that the upper and lower bounds depends on the asymptotic, and the approach are the same. – River Li Jun 14 '25 at 22:18

1 Answers1

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Maybe the following technique is better than induction.

$$a_{n+1}^3\geq a_n^3+3,$$ which by telescopic summation gives $$a_n^3\geq a_1^3+3(n-1)=3n-2$$ and $$a_n\geq\sqrt[3]{3n-2}.$$ Now, $$a_{n+1}^3=a_n^3+3+\frac{3}{a_n^3}+\frac{1}{a_n^6}\leq a_n^3+3+\frac{3}{3n-2}+\frac{1}{(3n-2)^2}.$$ Thus, $$a_{n+1}^3<1+3n+\int\limits_1^{n+1}\frac{1}{x-\frac{2}{3}}dx+\int\limits_1^{n+1}\frac{1}{(3x-2)^2}dx=$$ $$=1+3n+\ln\left(n+\frac{1}{3}\right)+\ln3-\frac{1}{3(3n+1)}+\frac{1}{3}<3n+3+\ln(n+1),$$ which says $$a_n<\sqrt[3]{3n+\ln{n}}.$$ By the way, easy to see that: $$\sqrt[3]{3n+\ln{n}}\leq \sqrt[3]{3n+\frac{31}{15}+\ln\frac{3n-1}5-\frac1{9n-3}}$$ for any natural $n.$

Michael Rozenberg
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    This isn't really what I was looking for. What if we are to determine $\left\lfloor10^{10}a_{900}\right\rfloor$? I wonder how we can derive the estimation of$$\sqrt[3]{3n+\frac{31}{15}+\ln\frac{3n-1}5-\frac1{9n-3}}$$and even perhaps, how to keep on improving it? – youthdoo Jun 14 '25 at 10:32
  • @youthdoo I think your question was to find $[a_{900}]$. No? Also, I think my estimation is better. – Michael Rozenberg Jun 14 '25 at 10:34
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    No, my question was not to find $[a_{900}]$, as I have already known how do do that. I am asking about what technique there is to estimate a recursive sequence in general. – youthdoo Jun 14 '25 at 10:36
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    @youthdoo I fixed. One of techniques, how we can get estimations, you saw. In the general your question is absurd. It's equivalent to the following question: what is a general way to solving any problems of math? – Michael Rozenberg Jun 14 '25 at 10:46
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    It has nothing to do with "any math problem". Estimating a recursive sequence is a much more specific task. – youthdoo Jun 14 '25 at 10:52
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    @youthdoo The point that Michael is making is that your question is so broad that it is impossible to answer. There are many different techniques depending on the exact recursive sequence you are working with. There is no such thing as a general answer. – FD_bfa Jun 14 '25 at 18:59
  • The person who proposed that upper bound presented his work. He used a method similar to yours, but dealt with some details differently. – youthdoo Jun 15 '25 at 02:06