Conjecture about the limit of $\left(\frac1n\sum_{k=r}^n n^{\frac1k}\right)^{n^{c}}$ By thinking a little further about my previous question, I made the following conjecture:
$$
\lim_{n\to\infty}{\left(\frac1n\sum_{k=r}^n n^{\frac1k}\right)^{n^{c}}}=\begin{cases}
\infty  & \text{if $c>\frac{r-1}{r}$} \\
e & \text{if $c=\frac{r-1}{r}$} \\
1 & \text{if $c<\frac{r-1}{r}$}
\end{cases}
$$
Where $r\in\mathbb{N}\setminus\{1\}$ and $c\in\mathbb{R}$. I approached it like in the question linked, but it didn't work. Is this conjecture true? And if so, how to find the right bounds to prove it?
 A: Roughly the sum in the power base is $n+n^{1/r}$. For large $k$ the terms are close to $1$; among the first "few" terms the very first one, $n^{1/r}$ determines the magniture order. So the base is approximately $1+n^{\frac1r-1}\approx e^{n^{\frac1r-1}}$.
In order to find a careful estimate, split the sum into 3 parts:
$$
\sum_{k=r}^n n^{\frac1k} 
= n^{\frac1r} + \sum_{r+1\le k\le\log n} n^{\frac1k} 
+ \sum_{\log n<k\le n} n^{\frac1k}.
$$
In the middle sum, estimate every term by $n^{\frac1{r+1}}$:
$$
\sum_{r+1\le k\le\log n} n^{\frac1k} = \mathcal{O}\left(n^{\frac1{r+1}}\log n \right).
$$
In the last sum, for $k>\log n$ we have
$$
n^{\frac1k} = e^{\frac{\log n}k} = 1 + \mathcal{O}\left(\frac{\log n}k\right),
$$
so
$$
\sum_{\log n<k\le n} n^{\frac1k}
= \sum_{r+1\le k\le\log n} \left( 1+\mathcal{O}\bigg(\frac{\log n}k\bigg) \right)
= n +\mathcal{O}(1) +\mathcal{O}\left(\sum_{k=r}^n \frac{\log n}k\right) 
= n +\mathcal{O}\big(\log^2n\big).
$$
Hence,
$$
\frac1n \sum_{k=r}^n n^{1/k} 
= \frac1n \left( n + n^{\frac1r}
  +\mathcal{O}\left(n^{\frac1{r+1}}\log n \right) \right)
= 1 + (1+\mathcal{o}(1))n^{\frac1r-1}
$$
so
$$
\log\left(\frac1n \sum_{k=r}^n n^{1/k}\right)^{n^c}
= n^c \cdot (1+\mathcal{o}(1))n^{\frac1r-1}
= (1+\mathcal{o}(1)) \cdot n^{c-\frac{r-1}r}.
\tag1
$$
If $c>\frac{r-1}r$, $c=\frac{r-1}r$ or $c<\frac{r-1}r$, then 
(1) tends to $\infty$, $1$ or $0$, respectively; the original sequence (without logarithm) tends to $\infty$, $e$ or $1$, respectively --- as you conjectured.
A: Since $1+x\le e^x\le1+\frac x{1-x}$, for $k\lt n-\log(n)$, we have that
$$
1+\frac{\log(n)}{n-k}\le n^{\frac1{n-k}}\le1+\frac{\log(n)}{n-k-\log(n)}\tag{1}
$$
Therefore, since $\sum\limits_{k=1}^{\lfloor n-log(n)\rfloor}\frac1k=O\left(\log(n)\right)$,
$$
\begin{align}
\small\frac1n\sum_{k=0}^{\lfloor n-\log(n)-1\rfloor}n^{\frac1{n-k}}
&\small=\frac{\lfloor n-\log(n)-1\rfloor}n+\frac{\log(n)}n\overbrace{\sum_{k=0}^{\lfloor n-\log(n)-1\rfloor}O\left(\frac1{n-k-\log(n)}\right)}^{O(\log(n))}\\
&\small=1+O\left(\frac{\log(n)^2}n\right)\tag{2}
\end{align}
$$
Furthermore, since $\log(n)\,n^{-\frac2{r(r+2)}}=o\left(n^{-\frac1{r(r+1)}}\right)$,
$$
\begin{align}
\frac1n\sum_{k=r}^{\lfloor\log(n)+1\rfloor}n^{\frac1{k}}
&=n^{-\frac{r-1}r}\left(\vphantom{n^{\frac12}}\right.1+n^{-\frac1{r(r+1)}}+\overbrace{n^{-\frac2{r(r+2)}}+n^{-\frac3{r(r+3)}}+\dots}^{\text{fewer than $\log(n)$ terms}}\left.\vphantom{n^{\frac12}}\right)\\
&=n^{-\frac{r-1}r}+O\left(n^{-\frac r{r+1}}\right)\tag{3}
\end{align}
$$
Adding $(2)$ and $(3)$, we get
$$
\begin{align}
\frac1n\sum_{k=r}^nn^{\frac1{k}}
&=1+n^{-\frac{r-1}r}+O\left(n^{-\frac r{r+1}}\right)\tag{4}
\end{align}
$$
and therefore, since $n^{-\frac r{r+1}}=o\left(n^{-\frac{r-1}r}\right)$,
$$
\begin{align}
\lim_{n\to\infty}\left(\frac1n\sum_{k=r}^nn^{\frac1{k}}\right)^{\large n^c}
&=\lim_{n\to\infty}\left(1+n^{-\frac{r-1}r}+O\left(n^{-\frac r{r+1}}\right)\right)^{\large n^c}\\
&=\left\{\begin{array}{}
1&\text{if }c\lt\frac{r-1}r\\
e&\text{if }c=\frac{r-1}r\\
\infty&\text{if }c\gt\frac{r-1}r
\end{array}\right.\tag{5}
\end{align}
$$
