# 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
2 years, 4 months ago

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

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

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

$$= \boxed{\gamma}$$

- 2 years, 3 months ago

exact intended solution!

upvoted :)

- 2 years, 3 months ago