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So I've been playing around with the following double integral $S$ where:

$$ S= \lim_{z \to \infty}\frac{1}{z^2} \int_{0^+}^z \exp\left({\frac{\int_{0^+}^y \ln (x) dx}{y}}\right) dy $$

Now, if I do the $dy$ integral I get:

$$ S =\lim_{z \to \infty}\frac{1}{z^2} \int_{0^+}^z \exp\left({\frac{y \ln y - y}{y}}\right) dy $$

OR:

$$ S =\lim_{z \to \infty}\frac{1}{z^2} \int_{0^+}^z \exp({{ \ln y - 1}}) dy $$

OR:

$$ S = \frac{e^{-1}}{2}$$

Let us use limit as a sum on the integrals on the $S$ double integral where $\epsilon \to 0$, $N \to \infty $ and $N \epsilon = z \to \infty$:

$$ S = \lim_{N \to \infty} \frac{\exp({ \frac{\ln (\epsilon)}{\epsilon} \cdot \epsilon}) \epsilon + \exp({ \frac{\ln (2! \epsilon^2)}{2\epsilon} \cdot \epsilon}) \epsilon + \dots + \exp({ \frac{\ln (N! \epsilon^N)}{N\epsilon} \cdot \epsilon}) \epsilon}{(N \epsilon)^2 }$$

$$ S = \lim_{N \to \infty} \frac{1 + (2!)^{1/2} + (3!)^{1/3} + (4!)^{1/4} + \dots + (N!)^{1/N}}{N^2} $$

Hence,

$$e^{-1}/ 2 = \lim_{N \to \infty} \frac{1 + (2!)^{1/2} + (3!)^{1/3} + (4!)^{1/4} + \dots + (N!)^{1/N}}{N^2} $$

Question

Is this correct? (I am worried about the use of limit of a sum twice without any concern of convergence).

Thomas Andrews
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More Anonymous
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  • Stirling’s approximation would indicate this might be true. Maybe some application of The Stolz-Cesaro Theorem. https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem – Thomas Andrews Nov 16 '21 at 14:57

1 Answers1

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The limit

$$ \lim_{N\to\infty} \frac{\sum_{k=1}^{N} (k!)^{1/k}}{N^2} = \frac{1}{2e} $$

is an immediate consequence of the Stolz–Cesàro theorem applied to the well-known limit

$$ \lim_{N\to\infty} \frac{(N!)^{1/N}}{N} = \frac{1}{e}. $$

(For example, see this answer for the proof of the limit above.)

Sangchul Lee
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