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