Let $(a_n)_{n\ge 1}$ be an increasing sequence of positive real numbers and suppose there is a constant $b > 0$ and $\varepsilon > 0$ such that
$$
a_{\lfloor n + \sqrt{n/2}\rfloor} \le a_n\left(1 + \frac{b}{n^{\frac{1}{2}+\varepsilon}}\right)
$$
for all $n$ sufficiently large. Does it follow that $a_n$ is bounded? i.e., is $\limsup a_n < \infty$?
If the left hand side of the inequality was $a_{n+1}$, then we could use induction to derive
$$
a_{n + 1} \le a_{n_0}\prod_{k=n_0}^{n}\left(1 + \frac{b}{k^{\frac{1}{2}+\varepsilon}}\right)
$$
and even then, it is not clear whether $a_n$ is bounded. Moreover, the form of the index on the left hand side of the inequality makes it difficult to apply this method. Is there a way to do this?
Any help or comments are welcome.
Best Answer
Write $f(n) = \bigl\lfloor n + \sqrt{n/2} \bigr\rfloor$ and consider the sequence $(n_k)_{k\geq 0}$ defined by
$$ n_0 \geq 2 \qquad\text{and}\qquad n_{k+1} = f(n_k). $$
Since $f(n) \geq n+1$ for $n \geq 2$, we find that $(n_k)_{k\geq 0}$ is strictly increasing. Moreover, for any sufficiently large $n$, we have
$$ \sqrt{f(n)} \geq \biggl( n + \frac{\sqrt{n}}{2} \biggr)^{1/2} = \sqrt{n} \biggl( 1 + \frac{1}{2\sqrt{n}} \biggr)^{1/2} \geq \sqrt{n} \biggl( 1 + \frac{1}{5\sqrt{n}} \biggr) = \sqrt{n} + \frac{1}{5}. \tag{*} $$
So by choosing $n_0$ to be sufficiently large, it follows that
$$ \sqrt{n_k} \geq \sqrt{n_0} + \tfrac{1}{5}k \quad \text{for all} \quad k \geq 0 $$
Then
$$ a_{n_k} \leq a_{n_0} \prod_{j=0}^{k-1} \biggl( 1 + \frac{b}{n_j^{0.5+\epsilon}} \biggr) \leq a_{n_0} \prod_{j=0}^{k-1} \biggl( 1 + \frac{b}{(\sqrt{n_0} + \frac{1}{5}j)^{1+2\epsilon}} \biggr) $$
and this upper bound converges as $k\to\infty$. Therefore $(a_n)$ is bounded above.
Addendum. In fact we can give a precise range of $n$ for which the inequality $\text{(*)}$ holds.
To begin with, the first step of $\text{(*)}$ holds precise when $ \left\lfloor \sqrt{\frac{n}{2}} \right\rfloor \geq \frac{\sqrt{n}}{2} $, which turns out to be true if and only if $n \geq 18$.
Next, it follows that $\sqrt{1+x} \geq 1 + \frac{2}{5}x$ if and only if $0 \leq x \leq \frac{5}{4}$. So the third step of $\text{(*)}$ holds for any $n$.
Combining altogether, $\text{(*)}$ holds for any $n \geq 18$ and we may take $n_0 = 18$.