4
$\begingroup$

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?

$\endgroup$
8
+100
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $n^{-\frac r{r+1}}=o\left(n^{-\frac{r-1}r}\right)$, $$ \begin{align} \lim_{n\to\infty}\left(\frac1n\sum_{k=r}^nn^{\frac1{k}}\right)^{\large n^c} &=\lim_{n\to\infty}\left(1+n^{-\frac{r-1}r}+O\left(n^{-\frac r{r+1}}\right)\right)^{\large n^c}\\ &=\left\{\begin{array}{} 1&\text{if }c\lt\frac{r-1}r\\ e&\text{if }c=\frac{r-1}r\\ \infty&\text{if }c\gt\frac{r-1}r \end{array}\right.\tag{5} \end{align} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.