Upper bound of $a_{n+1}=a_n + \frac{1}{a_n}$

Question

How can I prove:

If $$a_0=\alpha>0\quad and\quad a_{n+1}=a_n + \frac{1}{a_n}$$, then $$a_n^2<\alpha^2+2n+\frac{1}{\alpha^2}+\frac{1}{2}ln\left ( \frac{2n}{\alpha^2}+1 \right )$$ ?

I'll really appreciate your help. Thanks.

Answer 1

We have $\displaystyle a_{n+1}^2=a_n^2+\frac{1}{a_n^2}+2$. Hence for $n\geq 1$ $$a_n^2=a_0^2+2n+\sum_{k=0}^{n-1}\frac{1}{a_k^2}$$ This imply that $\displaystyle a_k^2\geq a_0^2+2k$ for $k\geq 1$. Hence $$a_n^2\leq a_0^2+2n+\frac{1}{a_0^2}+\sum_{k=1}^{n-1}\frac{1}{a_0^2+2k}$$

Now $$\sum_{k=1}^{n-1}\frac{1}{a_0^2+2k}\leq \int_0^{n}\frac{dt}{a_0^2+2t}=\frac{1}{2}\log (\frac{2}{a_0^2}n+1)$$ and we are done.

Answer 2

This is something I typed on October 31. It goes with an old question I can't find. It does indicate that both lower and upper bounds for $a_n$ can probably be proven by induction, and should be considered simultaneously. I will think about it.

An approach:

Suppose we have continuous $f(x) $ for $x > -1,$ with $f > 0,$ $f' < 0,$ and $$\lim_{x \rightarrow +\infty} f(x) = 0.$$

Take a sequence with, say, $x_0 = 0,$ after which $x_{n+1} = x_n + f(x_n).$ Then $x_n$ grows without bound.

This is a little counterintuitive. It is an exercise in some famous book.

I think the desired answer is this: it is an increasing sequence, assuming it has an upper bound, it has a least upper bound $L,$ which is also its limit. However, for some large $n,$ we have $x_n > L - \frac{f(L)}{2}, $ so that $x_{n+1} > L + \frac{f(L)}{2}, $ contradicting $L$ being an upper bound and the assumption that there is any upper bound.

My Proof: however small it might be, $f(1) > 0.$ As long as $x_n < 1,$ we get $x_{n+1} > x_n + f(1).$ Therefore, if we take $$ N_1 = \left\lceil \frac{1}{f(1)} \right\rceil, $$ we have $x_n \geq 1$ for $n \geq N_1.$

However small it might be, $f(2) > 0.$ As long as $x_n < 2,$ we get $x_{n+1} > x_n + f(2).$ Therefore, if we take $$ N_2 = \left\lceil \frac{1}{f(2)} \right\rceil, $$ we have $x_n \geq 2$ for $n \geq N_1 + N_2.$

Next $x_n \geq 3$ for $n \geq N_1 + N_2 + N_3.$

And so on, forever. $$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

In comparison, what about a sequence that really does converge, where we merely say $$ y_{n+1} > y_n. $$ For example, just take $$ y_n = 1 - \frac{1}{n}. $$ This has limit $1.$ We really could work out some continuous function $g$ such that $$g(y_n) = y_{n+1} - y_n = \frac{1}{n} - \frac{1}{n+1} = \frac{1}{n^2 + n}.$$ Which seems interesting, until we point out that continuity of $g$ demands $$ g(1) = 0. $$ That is, the strict positivity of $f$ is crucial in the result proved above.