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Gamma time!

\[ \large \gamma = \lim_{x\to1^+} \sum_{n=1}^\infty \left( \dfrac1{n^x} - \dfrac1{x^n} \right) \]

Let \( \gamma\) denote the Euler-Mascheroni constant. Prove the limit above.

This is a part of the set Formidable Series and Integrals

Note by Hummus A
11 months ago

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\( \displaystyle \lim_{x\to1^+} \sum_{n=1}^\infty \left( \dfrac1{n^x} - \dfrac1{x^n} \right) = \lim_{x\to1^+} \sum_{n=1}^\infty \left( \dfrac1{n^x} - \int_{n}^{n+1} \dfrac{\mathrm{d}y}{y^x} \right)\)

\(\displaystyle = \sum_{n=1}^\infty \left( \dfrac1{n} - \int_{n}^{n+1} \dfrac{\mathrm{d}y}{y} \right)\)

\(\displaystyle= \lim_{n \to \infty} \left( H_{n} - \log n \right)\)

\(= \boxed{\gamma} \) Ishan Singh · 11 months ago

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@Ishan Singh exact intended solution!

upvoted :) Hummus A · 11 months ago

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