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Suppose $s$ is not an integer, let $\lambda(s)=\min_{n≥0}|s+n|$. Show that $\sum\limits_{n=1}^{\infty}(\frac{1}{n+s}-\frac{1}{n})\ll\frac{1}{\lambda(s)}+\log(|s|+2)$.

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    What 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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    $\ll$ is an equivalent symbol for the big $O$.2011-10-03
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    You 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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    I 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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    Also under discussion at mathoverflow.2011-10-03
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    Here'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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    Also, 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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    I think I have got a proof. Thanks.2011-10-03

1 Answers 1

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It is worth noting that this is the main term of the Digamma Function, namely we have that $$\frac{\Gamma'}{\Gamma}(s)=\frac{-1}{s}-\gamma+\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{s+n}\right).$$

Here is a proof of the asymptotic. It is Theorem C.1 of the appendix in Montgomery and Vaughn's Multiplicative Number Theory.:

First $$\sum_{n=1}^M \left(\frac{1}{n}-\frac{1}{s+n}\right)=\log M +\gamma-\sum_{n=0}^M \frac{1}{n+s}.$$ By Euler MacLaurin summation on $\frac{1}{x+s}$ we have $$\sum_{n=0}^M \frac{1}{n+s}=\log(M+s)-\log s +\frac{1}{2s}+\frac{1}{2(s+M)}+O(|s|^{-2}).$$ Combining these and taking $M\rightarrow \infty$ we have

$$ \sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{s+n}\right)=\log s +\gamma +\frac{1}{2s}+O\left(\frac{1}{|s|^{2}}\right)$$

which is stronger then your desired result.

Remark: From here with the fact that $\frac{\Gamma'}{\Gamma}(s)=\frac{d}{ds}\log (\Gamma(s))$ we can deduce Stirlings Approximation.

Remark 2: The $\frac{1}{\lambda(s)}$ you have above comes from the $\frac{1}{2s}$. I believe that adding $2$ in the logarithm makes us no longer need the constant $\gamma$.

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    Thanks for your detailed explanation, Eric.2011-10-03