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I am looking for a simple proof that $\zeta(\alpha)/\zeta(\alpha+1) \to 1$ as $\alpha \to \infty$ (where $\zeta(\alpha)$ denotes the Riemann zeta function, $\zeta(\alpha) = \sum \limits_{n\geq 1} n^{-\alpha}$).

(Edit: Silly miscalculation by me, the terms do not go to 0 but in fact to 1, so the limit is easy to evaluate.) Since both terms go to 0, the limit isn't entirely trivial to evaluate. Also, since both terms are infinite sums of exponents I can't really use l'Hopital's rule without things becoming incredibly messy. Could anyone suggest an obvious way to show/see that the limit is 1 (as numerical evaluation would seem to indicate)?

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Actually, $\lim_{x\to\infty}\zeta(x)=1$. – Dejan Govc May 5 '12 at 9:37
up vote 3 down vote accepted

As Dejan Govc points out, $\zeta(\alpha) \to 1$ as $\alpha \to \infty$. So clearly $\zeta(\alpha)/\zeta(\alpha+1) \to 1$ as $\alpha \to \infty$.

More interestingly, $\lim\limits_{\alpha\to\infty}\displaystyle\frac{\zeta(\alpha)-1}{\zeta(\alpha+1)-1}$ is equal to $2$, not $1$. But this is also easy to prove by comparing the first terms of each series.

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Yes, stupid of me - I was for some reason thinking of a sum of increasing exponents of fixed $n$ in my head, rather than a sum of fixed exponents of increasing $n$. Thanks! – Spyam May 5 '12 at 12:10

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