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According to Wikipedia, the generalized Euler constants are $$\gamma_\alpha=\lim_{n\to \infty} \Big(\sum_{k=1}^n \frac{1}{k^\alpha} - \int_1^n \frac{1}{x^\alpha} \,dx \Big)$$ for $0<\alpha<1$. I am curious why we don't consider $\alpha >1$? For example, $$\lim_{n\to \infty} \Big(\sum_{k=1}^n \frac{1}{k^2} - \int_1^n \frac{1}{x^2} \,dx \Big) = -1 +\lim_{n\to \infty} \Big(\frac{1}{n}+ \sum_{k=1}^n \frac{1}{k^2} \Big) = \zeta(2)-1=\frac{\pi^2}{6}-1.$$ Or in general $$\lim_{n\to \infty} \Big(\sum_{k=1}^n \frac{1}{k^\alpha} - \int_1^n \frac{1}{x^\alpha} \,dx \Big) = \zeta(\alpha)-\frac{1}{\alpha-1}$$ for natural number $\alpha \geq 2$. Is it because they are easily describable in terms of the zeta function?

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There certainly isn't any mathematical reason why we cannot consider $\alpha > 1$, since in such a case, both the sum and the integral are finite, thus $\gamma_\alpha$ is well-defined for $\alpha > 1$. But that's kind of the issue: it is somewhat trivial or to put it more mildly, not so interesting to consider these cases. What makes the generalized $\gamma_\alpha$ interesting is precisely the case where $0 < \alpha < 1$, since on this interval, both the sum and integral are divergent, but their difference is finite in a way that is not trivial and may lead to more interesting mathematics.

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  • $\begingroup$ Thanks! This makes a lot of sense $\endgroup$ Dec 10, 2020 at 0:24

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