Does series convergence imply a convergence rate, if $a_{n+1} \le C a_n$ for some $C\in\mathbb{R}$?

Question

I have a positive sequence $a_n$, such that $|\sqrt{a_n} - \sqrt{a_{n+1}}|< c\sqrt{a_n}$. In particular the rate of increase is limited $$ \sqrt{a_{n+1}} = |\sqrt{a_{n+1}}| \le |\sqrt{a_n}| + |\sqrt{a_{n+1}}-\sqrt{a_n}| \le (1+c)\sqrt{a_n} $$ In other words I have $$a_{n+1} \le (1+c)^2 a_n.$$

Additionally I know the sequence is summable $$ \sum_{n=1}^\infty a_n <\infty. $$ Now this implies that $\liminf_n na_n = 0$ (according to Rate of convergence of summable sequence although I haven't proven that myself yet). Unfortunately summability alone is not sufficient for the rate $o(1/n$) as answers to the question above prove by counterexample.

These counterexamples usually involve rapidly decreasing sequences with infrequent elements of order $1/n$ forcing $\limsup_n na_n =1$. If we force our sequence to be monotonous, we actually get the desired rate (If $(a_n)\subset[0,\infty)$ is non-increasing and $\sum_{n=1}^\infty a_n<\infty$, then $\lim_{n\to\infty}{n a_n} = 0$). Now my sequence is not necessarily decreasing, but still disallows these big jumps. It therefore stands to reason, that this "almost monotonously" decreasing sequence might have a rate attached to it as well.

Any ideas?

Application

It is relatively straightforward to show for $L$-smooth functions $f$ (the gradient is $L$-Lipschitz continuous), that the gradient descent iteration $$ x_{n+1} = x_n - h\nabla f(x_n) $$ with $h<2/L$ leads to $$ \sum_{k=n}^m |\nabla f(x_k)|^2 \le \frac{f(x_n) - f(x_m)}{h(1-\frac{L}2h)}\le \frac{f(x_n) - \min_x f(x)}{h(1-\frac{L}2h)} $$ and thus that for finite global minima, the gradient squares are summable. It would be nice to get a rate of convergence for the gradients $\sqrt{a_n} = |\nabla f(x_n)|$, which due to $L$-smoothness satisfy $$ \bigl||\nabla f(x_{n+1})| - |\nabla f(x_n)|\bigr| \le |\nabla f(x_{n+1})-\nabla f(x_n)| \le L |x_{n+1}-x_n| = Lh|\nabla f(x_n)| $$

Answer 1

Inspired by https://math.stackexchange.com/a/4605/445105 I thought I could prove a rate, but it turns out there was a mistake in the proof. Now I believe that no such rate is possible. Here is my attempt:

First we define $$ a^*_n := \min_{n\le k \le f(n)} a_k \qquad q(n) := \frac{a_{f(n)}}{a^*_n} \ge 1 \qquad S_n:= \sum_{k=0}^n a_k $$ Since $S_{f(n)} - S_n \le S_\infty - S_n \to 0$ due to summability, we want to bound $a_{f(n)}$ in terms of it.

From $a_{n+1}\le C a_n$ follows by induction \begin{equation} \label{eq: lower bound on sequence} a_{f(n)-k} \ge a_{f(n)}C^{-k} \end{equation} By definition of $a^*_n$ there must exist $m < f(n)-n$ such that $$ a_{f(n)} C^{-m} \ge a^*_n \ge a_{f(n)}C^{-(m+1)}. $$ So $$ C^{m+1} \ge \frac{a_{f(n)}}{a^*_n} = q(n) \ge C^m $$ and therefore $$ m + 1\ge \frac{\ln(q(n))}{\ln(C)} \ge m. $$ Now we split up our sums, we know $$ \begin{aligned} S_{f(n)} - S_n &\ge \sum_{k=m+1}^{f(n)-n} a_{f(n)-k} + \sum_{k=0}^{m} a_{f(n)-k}\\ &\ge [f(n)-n - m] a^*_n + a_{f(n)}\underbrace{ \sum_{k=0}^{m} C^{-k}}_{ = \frac{1-C^{-(m+1)}}{1-C^{-1}} }\\ &\ge a_{f(n)} \Bigl[ \frac{f(n)-n-m}{q(n)} + \frac{1-\exp(-(m+1)\ln(C))}{1-C^{-1}} \Bigr]\\ &\ge a_{f(n)} \Bigl[ \frac{f(n)-n-\frac{\ln(q(n))}{\ln(C)}}{q(n)} + \frac{1-\exp(-\ln(q(n)))}{1-C^{-1}} \Bigr]\\ &\ge a_{f(n)} \Bigl[ \frac{f(n)-n-\frac{\ln(q(n))}{\ln(C)}}{q(n)} + \frac{q(n)-1}{(1-C^{-1})q(n)} \Bigr] \end{aligned} $$ Therefore for $\delta(n) = f(n) - n$ $$ a_{f(n)} \le \underbrace{[S_\infty - S_n]}_{o(1)} \frac{q(n)}{\delta(n) - \frac{\ln(q(n))}{\ln(C)} + \frac{q(n)-1}{1-C^{-1}}} $$ if you now use $q(n) \le C^{m+1}\le C^{\delta(n)}$ you could optimize over $\delta(n)$ in the hope of improving the rate of convergence over $o(1)$. But this does not yield any improvement.