Suppose $s$ is not an integer, let $\lambda(s)=\min_{n≥0}|s+n|$. Show that $\sum\limits_{n=1}^{\infty}(\frac{1}{n+s}-\frac{1}{n})\ll\frac{1}{\lambda(s)}+\log(|s|+2)$.
An estimate of a series
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analysis
asymptotics
analytic-number-theory
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0What is $\ll $ here? I think of it as much less, but if $s=0.1$, for example, it doesn't seem too much less to me (-0.1 vs 0.8) – 2011-10-03
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4$\ll$ is an equivalent symbol for the big $O$. – 2011-10-03
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0You probably meant $\sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+s} \right)$ in the left-hand-side to agree with asymptotic expansion for large positive values of $s$. – 2011-10-03
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0I think it can be reduced to showing that $\sum\limits_{n=1}^{|s| + 1}\frac{1}{n+s}\ll\frac{1}{\lambda(s)}+\log(|s|+2)$. – 2011-10-03
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0Also under discussion at mathoverflow. – 2011-10-03
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0Here's a plausible sketch. Let $$\phi(s)=\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+s}\right).$$ Fix $\alpha\in(0,1)$ and show that $\phi(m+\alpha)-\phi(\alpha)=O(\log m)$ in the integers in both $+\infty$ and $-\infty$ directions. Then fix $m\in\mathbb{Z}$ and let $\alpha$ vary in $(0,1)$; show $\phi(m+\alpha)=O(1/\alpha)$ as $\alpha\to0^+$ and $O(1/(1-\alpha))$ as $\alpha\to1^-$. The trick is to reason out actual, tangible constants in all of these big $O$ asymptotics, so that you can "glue" them together to apply validly over all of $\mathbb{R}$... – 2011-10-03
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0Also, note $$\lambda(s)=\begin{cases}s&s\ge0\\\min\{\mathrm{frac}(s),\mathrm{frac}(-s)\}&s\le0.\end{cases}$$ – 2011-10-03
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0I think I have got a proof. Thanks. – 2011-10-03