How to find range $a_{75}$ of the term of the series $a_n=a_{n-1}+ {1 \over {a_{n-1}}} $

Question

If $a_1=1$ and for n>1$$a_n=a_{n-1}+ {1 \over {a_{n-1}}} $$

$a_{75}$ lies between

(a) (12,15)

(b) (11,12)

(c) (15,18)

Now , in this question, I rewrote, $a_n-a_{n-1} = {1 \over {a_{n-1}}}$, to make a telescopic type and summed the terms to get

$$a_n-a_{1}= {1 \over {a_{n-1}}}+ {1 \over {a_{n-2}}}...{1 \over {a_{1}}}$$

And as ${1 \over {a_{r}}} <1$ for all r , I got $a_{75} <76$ , but it's of no help as , the upper bound is much higher.

I got no further ideas , on how to solve it, please help.

Answer 1

Batominovski gives an answer, but can we improve the upper bound?

Given the squared relation

$a_n^2=a_{n-1}^2+2+(1/a_{n-1})^2$

we know that the increments from $a^2_{n-1}$ to $a^2_n$ are greater than $2$, but by that very fact the additional increment beyond $2$ must also be less than or equal to $1/(2n-1)$.

So we have

$1+2(n-1)<a_n^2<l\le1+2(n-1)+\color{blue}{(1/1+1/3+1/5+...1/(2n-1))}$

where the series in blue is bounded by the harmonic series and is less than $\ln n +1$.

So

$1+2(n-1)<a_n^2\le 1+2(n-1)+\ln n+1$

and putting $n=75$ renders $a_{75}$ not only between $12$ and $15$, but more tightly between $12$ and $13$.

Also this proves that as $n\to\infty$, $a_n/\sqrt{n}\to \sqrt2$.

Answer 2

It is easy to see that $a_n\geq 1$ for every positive integer $n$, and the equality holds if and only if $n=1$. Note that $$a_n^2=\left(a_{n-1}+\frac{1}{a_{n-1}}\right)^2=a_{n-1}^2+2+\frac{1}{a_{n-1}^2}$$ for each integer $n\geq 2$. Since $0< \dfrac{1}{a_{n-1}^2}\leq 1$, we obtain $$a_{n-1}^2+2<a_n^2\leq a_{n-1}^2+3\,.$$ This shows that $$2n-1\leq a_n^2\leq 3n-2$$ for all positive integers $n$. The left-hand side inequality is an equality if and only if $n=1$. The right-hand side inequality is an equality if and only if $n=1$ or $n=2$. In particular, $$12^2<149<a_{75}^2<223<15^2\,.$$

You can easily show that $12<a_{75}<13$ in the same manner as follows. Since $a_2=2$, we can see that $$a_n^2\leq a_{n-1}^2+\frac{9}{4}$$ for all integers $n\geq 3$. Ergo, for every integer $n\geq 2$, we have $$a_n^2\leq \frac{9n-2}{4}\,,$$ with equality condition $n\in\{2,3\}$. Particularly, this shows that $$a_{75}^2<\frac{673}{4}=168\frac{1}{4}<13^2\,.$$

I believe that $$\lim_{n\to\infty}\,\frac{a_n}{\sqrt{2n}}=1\,.$$ In fact, this should also be true: $$\lim_{n\to\infty}\,\big(a_n-\sqrt{2n}\big)=0\,.$$ Oscar Lanzi proved these equalities in another answer. I expect, however, that $$a_n=\sqrt{2n}+\mathcal{O}\left(\frac{1}{\sqrt n}\right)\,.$$ Below is the plot of $\sqrt{n}\,\big(a_n-\sqrt{2n}\big)$ versus $n$ ($n$ runs from $1$ to $10^6$). (This conjecture is proven to be wrong by Oscar.)

Perhaps, the conjecture above is not quite correct. Maybe, this is a better claim: $$a_n=\sqrt{2n}+\mathcal{O}\left(\frac{\ln(n)}{\sqrt n}\right)\,.$$ Below is the plot of $\dfrac{\sqrt{n}}{\ln(n)}\,\big(a_n-\sqrt{2n}\big)$ versus $n$ ($n$ runs from $1$ to $10^6$). It appears from this plot that $$a_n\approx\sqrt{2n}+\frac{1}{4\sqrt2}\,\left(\frac{\ln(n)}{\sqrt n}\right)+\mathcal{O}\left(\frac1{\sqrt{n}}\right)\,.$$ (This conjecture is proven to be right by Oscar, and the asymptotic behavior above is true regardless of the initial value $a_1>0$.) This approximation is quite good (e.g., it says $a_{75}\approx 12.336$, while the actual value is $a_{75}\approx 12.324$).