A Limit with Fractions, Roots, Powers and Series

Problem

A Limit with Fractions, Roots, Powers and Series

Solution 1

Applying the Stolz-Cesaro theorem,

$\displaystyle \begin{align} \lim_{n\to \infty}\frac{1+\frac{1}{\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}}+\ldots+\frac{1}{\sqrt[3]{n}}}{n^{\frac{2}{3}}} &=\lim_{n\to \infty}\frac{\frac{1}{\sqrt[3]{n+1}}}{(n+1)^{\frac{2}{3}}-n^{\frac{2}{3}}}\\ &=\lim_{n\to \infty} \frac{1}{(n+1)^{\frac{1}{3}}\cdot n^{\frac{2}{3}}\left[\left(1+\frac{1}{n}\right)^{\frac{2}{3}}-1\right]}\\ &=\lim_{n\to \infty} \frac{1}{\left(1+\frac{1}{n}\right)^{\frac{1}{3}}\cdot \frac{\left(1+\frac{1}{n}\right)^{\frac{2}{3}}-1}{\frac{1}{n}}}\\ &=\frac{1}{\frac{2}{3}}=\frac{3}{2} \end{align}$

Similarly,

$\displaystyle \lim_{n\to \infty} \frac{1+\frac{1}{\sqrt[5]{2}}+\frac{1}{\sqrt[5]{3}}+\ldots+\frac{1}{\sqrt[5]{n}}}{n^{\frac{4}{5}}}=\frac{5}{4}.$

Thus

$\displaystyle \begin{align} \Omega&=\lim_{n\to \infty}\frac{\sqrt[3]{\bigg(1+\frac{1}{\sqrt[5]{2}}+\frac{1}{\sqrt[5]{3}}+\ldots+\frac{1}{\sqrt[5]{n}}}\bigg)^2}{\sqrt[5]{\bigg(1+\frac{1}{\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}}+\ldots+\frac{1}{\sqrt[3]{n}}\bigg)^4}}\\ &=\lim_{n\to \infty} \frac{\bigg(1+\frac{1}{\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}}+\ldots+\frac{1}{\sqrt[3]{n}}\bigg)^{-\frac{4}{5}}}{\bigg(1+\frac{1}{\sqrt[5]{2}}+\frac{1}{\sqrt[5]{3}}+\ldots+\frac{1}{\sqrt[5]{n}}\bigg)^{-\frac{2}{3}}}\\ &=\lim_{n\to \infty}\bigg(\frac{1+\frac{1}{\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}}+\ldots+\frac{1}{\sqrt[3]{n}}}{n^{\frac{2}{3}}}\bigg)^{-\frac{4}{5}}\cdot \bigg(\frac{n^{\frac{5}{4}}}{1+\frac{1}{\sqrt[5]{2}}+\frac{1}{\sqrt[5]{3}}+\ldots+\frac{1}{\sqrt[5]{n}}}\bigg)^{-\frac{2}{3}}\\ &=\frac{\bigg(\frac{5}{4}\bigg)^{-\frac{2}{3}}}{\bigg(\frac{3}{2}\bigg)^{-\frac{4}{5}}}=\frac{\bigg(\frac{3}{2}\bigg)^{\frac{4}{5}}}{\bigg(\frac{5}{4}\bigg)^{\frac{2}{3}}} \end{align}$

Solution 2

For some $k>1$, consider the series

$\displaystyle S_n=\frac{1}{1+\frac{1}{2^{1/k}}+\frac{1}{3^{1/k}}+...+\frac{1}{n^{1/k}}}.$

This series converges to $0$ as $n\rightarrow \infty$. We claim this series approaches $0$ with the leading term as $\displaystyle \frac{1}{n^q}$ for some $q\lt 1$. Thus, if we express $S_n$ as

$\displaystyle S_n=\frac{W_n}{n^q},$

then the series $W_n$ has a finite non-zero limit (say $x$) as $n\rightarrow \infty.$ Thus,

$\displaystyle \begin{align} \frac{1}{S_n}-\frac{1}{S_{n-1}}&=\frac{1}{n^{1/k}} \\ \frac{n^q}{W_n}-\frac{(n-1)^q}{W_{n-1}}&=\frac{1}{n^{1/k}} \end{align}$

In the limit as $n\rightarrow \infty$,

$\displaystyle \begin{align} x&=n^{1/k}[n^q-(n-1)^q] \\ &=n^{q+1/k}\left[1-\left(1-\frac{1}{n}\right)^q\right] \\ &\sim n^{q+1/k}\frac{q}{n} \\ &\sim qn^{q+1/k-1} \end{align}$

For this to be a constant, the leading power of $n$ has to be $0$. Thus, $\displaystyle q=1-\frac{1}{k}$ and $x=q$.

Thus upto the leading term,

$\displaystyle \begin{align} \frac{1}{1+\frac{1}{2^{1/3}}+\frac{1}{3^{1/3}}+\ldots+\frac{1}{n^{1/3}}}&\sim \frac{2}{3n^{2/3}}\\ \frac{1}{1+\frac{1}{2^{1/5}}+\frac{1}{3^{1/5}}+\ldots+\frac{1}{n^{1/5}}}&\sim \frac{4}{5n^{4/5}}, \end{align}$

and the limit for the original problem is

$\displaystyle L=\frac{(2/3)^{4/5}}{(4/5)^{2/3}}\sim 0.839.$

Solution 3

We first prove the following lemma:

If $\alpha \gt -1,\,$ then

$\displaystyle \lim_{n\to\infty}\left(\frac{1^{\alpha}+2^{\alpha}+\ldots+n^{\alpha}}{n^{\alpha +1}}\right)=\frac{1}{\alpha +1}.$

Indeed,

$\displaystyle \begin{align} \lim_{n\to\infty}\left(\frac{1^{\alpha}+2^{\alpha}+\ldots+n^{\alpha}}{n^{\alpha +1}}\right)&=\lim_{n\to\infty}\left(\frac{1}{n}\cdot\sum_{k=1}^n\left(\frac{k}{n}\right)^{\alpha}\right)\\ &=\int_0^1x^{\alpha}dx=\frac{1}{\alpha+1}. \end{align}$

Now, with $\displaystyle \alpha=-\frac{1}{5},\,$

$\displaystyle \lim_{n\to\infty}\frac{1+2^{-1/5}+\ldots+n^{-1/5}}{n^{4/5}}=\frac{5}{4}$

and with $\displaystyle \alpha=-\frac{1}{3},\,$

$\displaystyle \lim_{n\to\infty}\frac{1+2^{-1/3}+\ldots+n^{-1/3}}{n^{2/3}}=\frac{3}{2}$

Interchanging the operations,

$\displaystyle\begin{align}\lim_{n\to \infty} \frac{\displaystyle \left(\sum_{k=1}^n\frac{1}{\sqrt[5]{k}}\right)^{2/3}}{\displaystyle \left(\sum_{k=1}^n\frac{1}{\sqrt[3]{k}}\right)^{4/5}}&=\lim_{n\to \infty} \frac{\displaystyle \left( \left[\sum_{k=1}^n\frac{1}{\sqrt[5]{k}}\right]/n^{4/5}\right)^{2/3}}{\displaystyle \left(\left[ \sum_{k=1}^n\frac{1}{\sqrt[3]{k}}\right]/n^{2/3}\right)^{4/5}}\\ &=\frac{\displaystyle \left(\frac{3}{2}\right)^{4/5}}{\displaystyle \left(\frac{5}{4}\right)^{2/3}}. \end{align}$

Solution 4

$\displaystyle \begin{align} a_0&=\int_0^K \frac{1}{\sqrt[5]{u}} \, du=\frac{5 }{4}K^{4/5}\\ a_1&=\int_0^K \frac{1}{\sqrt[3]{u}} \, du=\frac{3 }{2}K^{2/3} \end{align}$

$\displaystyle \Omega \to \frac{a_0^{2/3}}{a_1^{4/5}}$ for $N$ large, which is what we are looking for. So, as $K$ comes out from $\frac{\displaystyle \left(\frac{5 K^{4/5}}{4}\right)^{2/3}}{\displaystyle \left(\frac{3 K^{2/3}}{2}\right)^{4/5}}$,

$\displaystyle \frac{a_0^{2/3}}{a_1^{4/5}}=\frac{5^{2/3}}{2^{8/15} 3^{4/5}}\simeq 0.838945.$

Added a Mathematica double check and intuition builder:

an illustration of a limit involving zeta functions

Acknowledgment

Dan Sitaru has kindly emailed me a LaTeX file with the above problem and his solution (Solution 1); the problem has been originally published at the Romanian Mathematical Magazine. Solution 2 is by Amit Itagi; Solution 3 is by Leo Giugiuc; Solution 4 is by N. N. Taleb.

 

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