# Bounding a Sum using Absoulte Convergence (Analytic Number Theory)

I'm reading the Multiplicative Number Theory lecture notes by Terence Tao, which can be found in his blog, and I stumbled into the following situation:

One can prove by estimating the sum of the function with its integral, then using the Euler product formula, and taking logarithms that for $s\in \mathbb{R}$, $s>1$ $$-\sum_{p}log(1-\frac{1}{p^s})=log(\frac{1}{1-s})+O(1-s).$$ Moreover, by taking the Taylor expansion, one can show that $$-log(1-\frac{1}{p^s})=\frac{1}{p^s}+O(\frac{1}{p^{2s}}).$$ From this, Tao states that because $\sum_{p}\frac{1}{p^{2s}}\leq \sum_{n}\frac{1}{n^2}$ is absolutely convergent, $$\sum_{p}\frac{1}{p^s}=log(\frac{1}{s-1})+O(1).$$ Why does the last equality hold? The definition Tao is using for big O notation is that $X=O(Y)$ if there exists a constant $C$, such that $|X(s)|\leq C|Y(s)|$ where the same constant suffices for all values of $s$.

As $s>1$, $$\sum_{p}\frac1{p^{2s}}<\sum_{p}\frac1{p^2}<\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6.$$ The $O(1/p^{2s})$ is a function $h_p(s)$ with a bound $$|h_p(s)|\le\frac{C}{p^{2s}}$$ where $C$ is independent of $p$. Then $$\left|\sum_p h_p(s)\right|\le C\frac{\pi^2}6$$ for all $s$, and this is the $O(1)$.