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It is well known that $$\displaystyle \lim_{n \to+\infty }\sum_{k=1}^{n}\frac{k^{n}}{n^{n}}=\frac{e}{e-1}.$$

But if you try to verify this by considering its "continuous" counterpart which is

$$\displaystyle \lim_{x \to+\infty }\int_{1}^{x}\frac{y^{x}}{x^{x}}dy$$you see that the result is equal to $1$. Why do we get a different result in this case? Can you find an explanation? Any help will be highly appreciated!!

Dávid Laczkó
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    Why should those limits be equal? – Martin R Jul 05 '22 at 08:19
  • Because you may consider an $f(y)=1$ on $[1,2)$ , $f(y)=2$ on$ [2,3)$ e.t.c. and then approximate it with a differentiable function. But it is rather intuition that leads us to believe that! –  Jul 05 '22 at 08:31
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    Apart from some exceptions, the integral is just an approximation for the sum. For example: $\pi^2/6 = \sum 1/n^2 \ne \int_1^\infty dx/x^2 = 1$. – Martin R Jul 05 '22 at 08:33
  • I agree! This example is very convincing! –  Jul 05 '22 at 08:35
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    See https://math.stackexchange.com/a/164083/42969 for a way to calculate the limit of the sum with the help of integrals. – Martin R Jul 05 '22 at 08:48
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    The point is that the sum gives $$\lim_{n \to\infty }\sum_{k=1}^{n}\frac{k^{n}}{n^{n}}=\lim_{n\to\infty}\sum_{l=0}^{\sqrt n}e^{-l}=\sum_{l=0}^\infty e^{-l}$$ and the integral, in turn, $$\lim_{x \to\infty }\int_{1}^{x}\frac{y^{x}}{x^{x}}dy=\lim_{x \to\infty }\int_0^\sqrt x e^{-z}dz=\int_0^\infty e^{-z}dz$$ so, effectively, you compare $\sum_{l=0}^\infty e^{-l}$ and $\int_0^\infty e^{-z}dz$ – Svyatoslav Jul 05 '22 at 08:59

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