Putnam 2006 - Exercise B.6 - Alternative solution verification and questions about generalizations

CONTEXT

Here is the orginal problem statement.

Let $k$ be an integer greater than 1. Suppose $a_0 > 0$, and define $$ a_{n+1} = a_n + \frac{1}{\sqrt[k]{a_n}} $$ for $n > 0$. Evaluate $$ \lim_{n \to \infty} \frac{a_n^{k+1}}{n^k}. $$**

PROPOSED SOLUTION

1. Note that for $n>0$ $$a_n \geq a_0 + \frac1{\sqrt[k]{a_0}}\sum_{j=1}^n \frac1{\sqrt[k]{j}},$$ and hence $(a_n)$ diverges.

EDIT. As pointed out in comment, this step is not correct. I am working on an alternative approach.

EDIT 2. A nice approach is proposed in the comment by Thomas Andrews

2. Define $b_n = a_n^{\frac{k+1}k}$. (Observe that $(b_n)$, too, is divergent.) Now, we have \begin{eqnarray}\lim_{n\to \infty}(b_{n+1}-b_n)&= &\lim_{n\to \infty}\left(a_n+a_n^{-\frac1{k}}\right)^{\frac{k+1}k}-a_n^{\frac{k+1}k}=\\&=&\lim_{n\to\infty} \frac{\left(1+a_n^{-\frac{k+1}k}\right)^{\frac{k+1}k}-1}{a_n^{-\frac{k+1}k}}=\frac{k+1}{k},\end{eqnarray}where the fundamental limit$$\frac{(1+\alpha)^m-1}{\alpha}\to m, \ \ \ \mbox{for} \ \ \alpha\to 0$$has been used.
3. By Stolz-Cesàro Theorem, we have $$\lim_{n\to\infty}\frac{b_n}{n} = \frac{k+1}k,$$ and hence $$\lim_{n\to\infty} \frac{a_n^{k+1}}{n^k} = \left(\frac{k+1}k\right)^k.$$ $\blacksquare$

QUESTIONS

1. Is my solution correct?
2. Is it correct to state that the solution is valid also for any real number $k\geq 1$?
3. Can we further generalize? For example: what can be stated for $0<k<1$, maybe with some restriction on $a_0$?

Step (1) was a bit hand-wavy. The rest of the argument looks fine.

A replacement for step 1

Since $a_n$ is increasing, it must converge or be divergent to $+\infty.$

If $a_n$ converges to $a>0,$ then $$\sum_{n=0}^{\infty}(a_{n+1}-a_n)$$ converges to $a-a_0,$ and thus $1/a_n^{1/k}=a_{n+1}-a_n\to 0.$ But $1/a_n^{1/k}\to\frac1{a^{1/k}}\neq 0,$ which is a contradiction.

So $a_n$ diverges to $+\infty.$

This works for any sequence $$c_0>0, c_{n+1}=f(c_n)$$ where $f:\mathbb R^{+}\to\mathbb R^+$ is continuous, with $f(x)>x$ for all $x,$ by the same reasoning. We can always conclude that $c_n$ is divergent to $+\infty.$ Essentially, a limit would have to be a fixed point of $f,$ and since $c_n$ is increasing, divergence means divergence to $+\infty.$

It seems like it will work for all real $k>0$ and $k<-1.$ It won’t work when $k\in(-1,0)$ because $\frac{k+1}{k}<0,$ so $b_n$ doesn’t diverge then.

Generalization of part (2).

Let $g(x)$ be a positive increasing differentiable function with $g(x)\to\infty$ $$\lim_{x\to\infty} \frac{xg’(x)}{g(x)}=L.$$

If $$a_0>0,a_{n+1}=a_{n}\left(1+1/g(a_n)\right)$$ and $b_{n}=g(a_n).$ then I believe we can show: $$b_{n+1}-b_{n}\to L.$$

Specifically, $$g(a_{n+1})-g(a_n)=g(a_{n}+a_{n}/g(a_n)-g(a_n)=\frac{g’(z_n)a_n}{g(a_n)}$$ for some $z_n\in (a_n,a_{n+1}).$ So you need some tighter bound here on the rate of change on $f’$ to get the limit $L.$

If so, $\frac {b_{n}}n\to L,$ and hence $$\frac{g(a_n)}{n}\to L.$$

When $g(x)=x^{\alpha},\alpha>0,$ then $$\frac{xg’(x)}{g(x)}\to \alpha.$$ If $1-\alpha=-1/k,$ then $\alpha=\frac{k+1}{k}.$

Not sure about the conditions on $g.$ You might need something additional. Like $g’$ monotonic and $$\frac{g’(x)}{g’(x+x/g(x))}\to 1.$$ That might be expressed as some condition on $g’’,$ perhaps $$\frac{g’’(x)x}{g(x)g’(x)}\to 0.$$