Closed form for the sequence defined by $a_0=1$ and $a_{n+1} = a_n + a_n^{-1}$

Question

Today, we had a math class, where we had to show, that $a_{100} > 14$ for

$$a_0 = 1;\qquad a_{n+1} = a_n + a_n^{-1}$$

Apart from this task, I asked myself: Is there a closed form for this sequence? Since I didn't find an answer by myself, can somebody tell me, whether such a closed form exists, and if yes what it is?

Answer 1

A closed form I doubt there is. But asymptotics are easy: $$ a_{n+1}^2=a_n^2+2+1/a_n^2, $$ hence, for every $n\ge1$, $$ a_n^2=2n+1+\sum_{k=0}^{n-1}\frac1{a_k^2}.\qquad\qquad\qquad\qquad (*) $$ This shows that $a_n^2\ge2n+2$ for every $n\ge1$, for example $a_{100}\ge\sqrt{202}>10\sqrt{2}>14$. In particular, $a_n\to+\infty$. Plugging this into $(*)$ yields $a_n^2=2n+1+o(n)$ hence $$ \sqrt{2n}\le a_n\le\sqrt{2n}+o(\sqrt{n}). $$ At this point, we know that $a_n^2\ge2n+2$ for every $n\ge1$. Using $(*)$ again, one sees that, for every $n\ge1$, $$ a_n^2\le2n+2+\sum_{k=1}^{n-1}\frac1{2+2k}\le2n+2+\frac12\log(n). $$ Which already shows that $$14.2<a_{100}<14.3$$ Plugging this upper bound of $a_n^2$ into $(*)$ would yield a refined lower bound of $a_n^2$. And one could then plug this refined lower bound into $(*)$ again to get a refined upper bound. And so on, back and forth between better and better upper bounds and better and better lower bounds. (No more asymptotics here.)

Answer 2

I agree, a closed form is very unlikely. As for more precise asymptotics, I think $a_n = \sqrt{2n} + 1/8\,{\frac {\sqrt {2}\ln \left( n \right) }{\sqrt {n}}}-{\frac {1}{ 128}}\,{\frac {\sqrt {2} \left( \ln \left( n \right) -2 \right) ^{2} + o(1)} {{n}^{3/2}}}$

Answer 3

To elaborate on how I got my answer: I started with @Did's $a(n) \approx \sqrt{2n}$ and looked for a next term. $a(n) = \sqrt{2n}$ would make $ a(n+1) - (a(n) + a(n)^{-1}) = \sqrt {2\,n+2}-\sqrt {2}\sqrt {n}-1/2\,{\frac {\sqrt {2}}{\sqrt {n}}} = - \frac{\sqrt{2}}{8} n^{-3/2} + O(n^{-5/2})$. With $a(n) = \sqrt{2n} + c n^{-1/2}$ I don't get a change in the $n^{-3/2}$ term, so I tried $a(n) = \sqrt{2n} + c \ln(n) n^{-1/2}$ and got $a(n+1) - (a(n) + a(n)^{-1}) = (-\frac{2}{\sqrt{8}} + c) n^{-3/2} + \ldots$. So to get rid of the $n^{-3/2}$ term I want $c = \frac{2}{\sqrt{8}}$. Then look at the leading term for $a(n) = \sqrt{2n} + \frac{2}{\sqrt{8}} \ln(n) n^{-1/2}$ and continue in that vein...

Answer 4

From my answer here: Given $a_{1}=1, \ a_{n+1}=a_{n}+\frac{1}{a_{n}}$, find $\lim \limits_{n\to\infty}\frac{a_{n}}{n}$

Reposting here, as it is kind of lost in that thread and this thread is more suitable for it.

Note: I have no clue if a closed form exists, but here is an asymptotic estimate...

I think we can show that $$\displaystyle a_{n}^2 \sim 2n + \dfrac{\log n}{2} - C$$ for some constant $\displaystyle C \gt 0$

By $\displaystyle x_n \sim y_n$ I mean $\displaystyle \lim_{n \to \infty} (x_n - y_n) = 0$

Consider $b_n = a_{n}^2 - 2n$

Then we have that $\displaystyle b_{n+1} = b_n + \dfrac{1}{b_n + 2n}$

Notice that $b_0 \gt 0$ and thus $\displaystyle b_n \gt 0$.

(Note that the other thread linked above starts with $a_1 = 1$ and not $a_0 = 1$.)

We can easily show that $b_n \lt 2 + \log n$, as

$b_{n+1} - b_n = \dfrac{1}{b_n + 2n} \lt \dfrac{1}{2n}$

Adding up gives us the easy upper bound. Note, even though we can give tighter bounds, this is sufficient for our purposes.

Now we have that, for sufficiently large $\displaystyle m,n$

$\displaystyle b_{m+1} - b_n = \sum_{k=n}^{m} \dfrac{1}{b_k + 2k}$

we have that

$\displaystyle \sum_{k=n}^{m} \dfrac{1}{2k} \gt b_{m+1} - b_n \gt \sum_{k=n}^{m} \dfrac{1}{2k}(1- \dfrac{b_k}{2k})$

(Here we used $\displaystyle \dfrac{1}{1+x} \gt \ \ 1-x, 1 \gt x \gt 0$)

Now Since $b_k \lt 2 + \log k$, we have that

$\displaystyle \sum_{k=n}^{m} \dfrac{1}{2k} \gt b_{m+1} - b_n \gt \sum_{k=n}^{m} \dfrac{1}{2k} - \sum_{k=n}^{m} \dfrac{2 + \log k }{4k^2}$

Using the fact that $\displaystyle H_m - H_n = \log(\dfrac{m+1}{n}) + O(\dfrac{1}{n}) + O(\dfrac{1}{n} - \dfrac{1}{m})$, where $\displaystyle H_n = \sum_{k=1}^{n} \dfrac{1}{k}$ is the $\displaystyle n^{th}$ harmonic number.

We see that,

if $c_n = b_n - \dfrac{\log n}{2}$, then

$\displaystyle O(\dfrac{1}{n} -\dfrac{1}{m}) + O(\dfrac{1}{n}) \gt c_{m+1} - c_n \gt O(\dfrac{1}{n} -\dfrac{1}{m}) + O(\dfrac{1}{n}) -\sum_{k=n}^{m} \dfrac{2 + \log k }{4k^2}$

Now $\displaystyle \sum_{k=1}^{\infty} \dfrac{2 + \log k}{k^2}$ is convergent and so by the Cauchy convergence criteria, we have that $\displaystyle c_n$ is convergent.

Thus the sequence $\displaystyle a_{n}^2 - 2n - \dfrac{\log n}{2}$ converges and hence, for some $\displaystyle C$ we have that

$$\displaystyle a_{n}^2 \sim 2n + \dfrac{\log n}{2} - C$$

or in other words

$$\displaystyle a_{n} \sim \sqrt{2n + \dfrac{\log n}{2} - C}$$

A quick (possibly incorrect) computer simulation seems to show a very slow convergence to $\displaystyle C = 1.47812676429749\dots$

Note: Didier suggested an alternate proof in the comments below, which might simpler.

Answer 5

Let us consider the functional formulation. Given $y(0)=1$, $y' = \frac{1}{y}$ yields $ y(x) = \sqrt{2x+1}$.

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I am not saying $a(n)=y(n)$. Yet there is a link between the two approaches (finite difference).

Answer 6

Let $b_n=a_n^{-1}$, so that $b_{n+1}=(b_n+b_n^{-1})^{-1}$ is the $n$th iterate of $$f(x)=(x+x^{-1})^{-1}=x-x^3+x^5-x^7+\cdots$$ The correct asymptotics have been given elsewhere, but this answer is to point out that there is a general method applicable to any function $f$ analytic at 0 with a series expansion $x+a_kx^k+a_{k+1}x^{k+1}+\cdots$ with $k>1$ and $a_k<0$: I learned about it in Chapter 8 of de Bruijn's Asymptotic Methods in Analysis, where it is shown that if the starting value is small enough, the $n$th iterate of $f$ is asymptotically equivalent to $((1-k)a_kn)^{-1/(k-1)}$. In our case, $k=3$ and $a_k=-1$ so $b_n \sim (2n)^{-1/2}$ and $a_n \sim (2n)^{1/2}$.

A sketch of the proof in the case at hand is as follows. First, show by induction that if $u_n$ is any sequence tending to zero and such that $$u_{n+1}=u_n-u_n^2+O(u_n^3)$$ then $u_n = n^{-1} + O(n^{-2}\log n)$. This is the tricky part of the proof.

Next make a substitution $z_n= 2b_n^2$. We have $$ b_{n+1}=b_n(1-b_n^2+b_n^4-\cdots)$$ so $$z_{n+1}=z_n\left(1-\frac{1}{2}z_n + \frac{1}{4}z_n^2+ \cdots\right)^2 = z_n-z_n^2 + \frac{3}{4}z_n^3 + \cdots$$ from which we get $z_n = n^{-1}+O(n^{-2}\log n)$ and the asymptotics for $b_n$ follow.

Incidentally, the sequence $b_n$ is what you get by using Newton's method to find a root of $xe^{-x^{-2}/2}$. But I couldn't find a way to make that shed any light on the asymptotics...