Does $ a_{\lfloor n + \sqrt{n/2}\rfloor} \le a_n\left(1 + \frac{b}{n^{\frac{1}{2}+\varepsilon}}\right) $ imply that $a_n$ is bounded? 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.
 A: As the sequence $(a_n)_n$ increases, it tends either to a finite limit (the sequence converges), either to $+\infty$ (the sequence diverges).
We will prove that the second case can't happen. Suppose by contradiction that the sequence tends to $+\infty$, then, there exists $n_0$ such that $a_{n_0} >2$.
Let us define the two sequence $g(k)$ and $f(k)$ with $k\in\Bbb N^*$ and $x>0$ as follows
$$
\begin{cases}
f(k) = \left[f(k-1)+\sqrt{\frac{f(k-1)}{2}}\right] \\
 f(0) = a_{n_0}
\end{cases}
$$
$$
\begin{cases}
g(k) = g(k-1)+\sqrt{\frac{g(k-1)}{2}}-1 \\
 g(0) = a_{n_0}
\end{cases}  \tag{1}
$$
It's easy to prove that the two sequences $f$ and $g$ are increasing and  $g(k) \le f(k)$. So,
$$a_{f(n)} \le a_0\prod_{k=1}^n\left( 1+ \frac{b}{(f(k))^{\frac{1}{2}+\epsilon}}\right) \le a_{n_0} \prod_{k=1}^n\left( 1+ \frac{b}{(g(k))^{\frac{1}{2}+\epsilon}}\right) $$
$$\implies \ln a_{f(n)} \le \ln a_{n_0} \sum_{k=1}^n\ln\left( 1+ \frac{b}{(g(k))^{\frac{1}{2}+\epsilon}}\right) \tag{2}$$
Let us study the sequence $g(k)$, it's easy to notice that $g(k) \to +\infty$ with $a_{n_0}>2$.
$$
\begin{align}
\sqrt{g(k)}-\sqrt{g(k-1)} &= \sqrt{g(k-1)} \left(\left(1+ \frac{1}{\sqrt{2g(k-1)}}-\frac{1}{g(k-1)}  \right)^{\frac{1}{2}}  -1\right) \\
& \approx  \sqrt{g(k-1)} \left(1+ \frac{1}{2} \frac{1}{\sqrt{2g(k-1)}}+\mathcal{O}\left(\frac{1}{g(k)}\right)    -1\right)\\
& \approx  \frac{1}{2\sqrt{2}}+\mathcal{O}\left(\frac{1}{\sqrt{g(k)}}\right)  \\
&\to \frac{1}{2\sqrt{2}}
\end{align}
$$
Applying the Stolz–Cesàro theorem, we deduce that $g(k)  \simeq \frac{k^2}{8} $ for $k \to +\infty$.
Return back to $(2)$, we have le term
$$
\begin{align}
\ln \left( 1+ \frac{b}{(g(k))^{\frac{1}{2}+\epsilon}}\right) \simeq \frac{b. 8 ^{\frac{1}{2}+\epsilon}}{k^{1+2\epsilon}}
\end{align}
$$
As the sum $\sum_{k=1}^{+\infty}\frac{1}{k^{1+2\epsilon}}$ converges to a limit $l$, we have the sequence $(a_{f(k)})_{k\ge 1}$ bounded. And so the sequence $(a_{n})_{n\ge 1}$ is also bounded (for each $n$, there exists a $k$ such that $f(k)>n$).
This result contradicts to what we supposed that the sequence tends to $+\infty$.
Conclusion:
we can conclude that the sequence $(a_n)_{n\ge 1}$ converges. Q.E.D
A: For any $n \ge 40$, consider the sequence  $$b_0=n, b_{k+1}= b_k+[\sqrt{b_k/2}]$$ Clearly, $b_k \ge n$ and $$b_{k+1} \ge b_k+\sqrt{n/2}-1$$ for all $k$. Hence,
$$b_{2 [\sqrt{n}]} \ge b_0+2 [\sqrt{n}]( \sqrt{n/2}-1) \stackrel{ n\ \ge 40}{\ge} b_0+n=2n$$
Besides, the condition on $(a_n)$ tells us that:
$$a_{b_{k+1}} \le a_{b_k}\left(1+\frac{b}{ (b_k)^{1/2+\epsilon}} \right) \le a_{b_k}\left(1+\frac{b}{n^{1/2+\epsilon}} \right) \le a_{b_k} \exp\left( \frac{b}{n^{1/2+\epsilon}}\right)$$
So
$$a_{2n} \stackrel{(a_n)\text{ increasing}}{\le}a_{b_{2[\sqrt{n}]}} \le a_{b_0} \exp\left( \frac{2[\sqrt{n}]b}{n^{1/2+\epsilon}}\right) \le a_n \exp(4bn^{-\epsilon})$$
Or,
$$a_{2n} \le a_n\exp( 4bn^{-\epsilon})$$
This means,
$$a_{2^{k+1}}\le a_{2^k}\exp( 4b2^{-k\epsilon})$$
for all $k \ge 6$. Or the sequence $( a_{2^k}, k \in \mathbb{N})$ is bounded.
Hence the conclusion.
A: 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 $n_0 = 18$.
