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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) –  Ross Millikan Oct 3 '11 at 4:30
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$\ll$ is an equivalent symbol for the big $O$. –  ksj03 Oct 3 '11 at 4:33
    
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 $s$. –  Sasha Oct 3 '11 at 5:20
    
I think it can be reduced to showing that $\sum\limits_{n=1}^{|s| + 1}\frac{1}{n+s}\ll\frac{1}{\lambda(s)}+\log(|s|+2)$. –  ksj03 Oct 3 '11 at 5:35
    
Also under discussion at mathoverflow. –  Gerry Myerson Oct 3 '11 at 6:30

1 Answer 1

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. –  ksj03 Oct 3 '11 at 12:09

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