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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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
1 Answers
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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0Thanks for your detailed explanation, Eric. – 2011-10-03