Question
Compute the products:
- $\prod_\limits{0<i<j<\infty} \left\{i^{\frac{1}{i}} - j^{\frac{1}{j}}\right\}$
- $\prod_\limits{0<i<j<2020} \left\{i^{\frac{1}{i}} - j^{\frac{1}{j}}\right\}$
From Isichenko (2021).
Attempt at solution
$\prod_\limits{0<i<j<\infty} \left\{i^{\frac{1}{i}} - j^{\frac{1}{j}}\right\} = \prod_\limits{i=0}^\infty \prod_\limits{j=i+1}^\infty \left\{i^{\frac{1}{i}} - j^{\frac{1}{j}}\right\}$
As $n$ tends towards infinity the $n$th root of n tends to 1.
$\lim_\limits{n \to \infty} \left\{\sqrt[n]{n}\right\} = \lim_\limits{n \to \infty} \left\{n^\frac{1}{n}\right\} = 1$
and
$i^\frac{1}{i} \gt j^\frac{1}{j} \; \forall \; j\gt i$
For every term in the product, since $j>i$ for every $i$
$\lim_\limits{i,j \to \infty} \left\{i^{\frac{1}{i}} - j^{\frac{1}{j}}\right\} = 0^+$
Therefore $\prod_\limits{0<i<j<\infty} \left\{i^{\frac{1}{i}} - j^{\frac{1}{j}}\right\} = 0$
For part 2. the same logic applies in that the term of the double product is always less than 1 and approaches 0. $\prod_\limits{0<i<j<2020} \left\{i^{\frac{1}{i}} - j^{\frac{1}{j}}\right\} = 0$ or very small.
Is my solution correct, and if so, is there a more formal way of proving this result? Is there an exact answer to part 2.?
Reference
Isichenko, M. (2021) Quantitative Portfolio Management: The Art and Science of Statistical Arbitrage. Wiley.