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Maybe not, but still I am interested by the following limits motivated by this question.

Let $$f(n) := \sum_{n+1}^{\infty} \frac{1}{k^2}.$$

I am interested in the limits

$$L(\alpha) := \lim_{n \rightarrow \infty} \frac{f(\alpha n )}{f(n)} $$

for numbers $\alpha \in \{2,3,4...\}$.

I have tried using the estimates on this page:

The sum of reciprocal squares: estimating the remainder

I have managed to show, for instance, that $ L(2) \leq \frac{3}{4}$ and that $L(3) \leq \frac{7}{9} $. But I have not even managed to prove that the limits are not $0$. (Maybe they are...)

Community
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Frank
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1 Answers1

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Since: $$ \frac{1}{k^2}=\left(\frac{1}{k}-\frac{1}{k+1}\right)+O\left(\frac{1}{k^3}\right) $$ it follows that: $$ f(n)=\sum_{k=n+1}^{+\infty}\frac{1}{k^2}=\frac{1}{n}+O\left(\frac{1}{n^2}\right) $$ hence for any positive $\alpha$ it happens that: $$ \lim_{n\to +\infty}\frac{f(\alpha n)}{f(n)}=\frac{1}{\alpha}.$$

Jack D'Aurizio
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    Aha. So there is regular variation with index $-1$...! Thanks a lot! – Frank Mar 05 '15 at 15:00
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    This is very elegant ! Just as an animal, I performed numerical simulation to get an idea but you came too fast for the old man. Thanks & cheers :-) – Claude Leibovici Mar 05 '15 at 15:27