Let $0 < \delta < r < 1$, where $r$ is fixed and $\delta $ not fixed, and let $f$ be a function defined on the integers. Assume further that for every $\delta$ there is some constant $0 < c_\delta < \infty$
$$ \frac{f(n)}{n^{(r-\delta)^{-1}}} \longrightarrow c_\delta, \quad \text{as } n \to \infty.$$
I want to show that: $f(n) = O (n^{1/r +\delta})$ for all $\delta > 0$.
I think (please correct me, if I am wrong), that from the convergence above we get $f(n) = O(n^{(r-\delta)^{-1}})$ for all $0 < \delta< r$.
Edit: Ok, here my try: Claim: $f(n) = O(n^{(r-\delta)^{-1}})$ for all $0 < \delta < r$.
Let $\epsilon >0$. $\forall \delta > 0 \ \exists N_\delta$ s.t. $ \forall n \geq N_\delta:$ $\vert \frac{f(n)}{n^{(r-\delta)^{-1}}} - c_\delta \vert < \epsilon$. Then, set $M := \max_\delta \{c_\delta\} + \epsilon$ which satisfies $f(n) \leq M n^{(r-\delta)^{-1}}$ for all $n \geq \max_\delta \{ N_\delta\}$. This should prove the claim. Correct?
The claim is not true. Take $r=1/2$ and $\delta=1/4$ and $f(n) = n^{(r-\delta)^{-1}} = n^4$. Then $n^{\frac{1}{r} + \delta} = n^{2.25}$ but $n^4 \notin O(n^{2.25})$.
More generally, the issue is that $\frac{1}{r-\delta} > \frac{1}{r} + \delta$ for $\delta > 0$, so $n^{(r-\delta)^{-1}}$ grows much faster than $n^{\frac{1}{r} + \delta}$.