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I must evaluate: $$\lim_{n\to\infty} \int_0^1 \frac{1}{\lfloor \frac{1}{x} \rfloor ^n}\text{d}x$$ As far as I've done, substituting $1/x=t$ it is: $$\lim_{n\to\infty} \int_0^1 \frac{1}{\lfloor \frac{1}{x} \rfloor ^n}\text{d}x=\lim_{n\to\infty} \int_1^\infty \frac{1}{\lfloor t \rfloor ^n t^2}\text{d}t$$ $$=\lim_{n\to\infty} \sum_{k=1}^\infty\int_k^{k+1}\frac{1}{\lfloor t \rfloor ^n t^2}\text{d}t=\lim_{n\to\infty}\sum_{k=1}^\infty\int_k^{k+1}\frac{1}{k^nt^2}\text{d}t$$ $$=\lim_{n\to\infty} \sum_{k=1}^\infty \frac{1}{k^n}\left(\frac{1}{k}-\frac{1}{k+1}\right)$$ and I got stuck, because I don't know how to evaluate the latter sum. I tried Cauchy product for series because of the telescopic term $1/k-1/(k+1)$, but to me it didn't seem to work. This seems to converge to $1/2$, but I couldn't prove it. I strongly prefer hints to a complete answer, thanks.

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Hint: the sum from $k=2$ to $\infty$ converges to $0.$

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    $\begingroup$ Thank you for the help, maybe I got it. It is $0 \le \sum_{k=2}^\infty \frac{1}{k^{n}}\left(\frac{1}{k}-\frac{1}{k+1}\right) \le \frac{1}{2^{n}} \sum_{k=2}^\infty \left(\frac{1}{k}-\frac{1}{k+1}\right)=\frac{1}{2^{n+1}} \to 0$ as $n \to \infty$. Correct? $\endgroup$
    – Bernkastel
    Nov 25, 2022 at 22:25
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    $\begingroup$ Yes! You got it. $\endgroup$ Nov 25, 2022 at 22:28

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