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This is a step in Titchmarsh's The theory of Riemann Zeta functions pg 2 of Chpt 1 eq 1.13 which I do not find obvious, though I can prove it directly. Assume $\sigma=Re(s)>1$.

Denote by $\pi(x)$ the number of primes $\leq x$. Consider the following summation which is basically $-\log(\zeta(s))$.

\begin{align} \sum_{n\geq 2}a_n &:=\sum_{n\geq 2}(\pi(n)-\pi(n-1))\log\left(1-\tfrac{1}{n^s}\right) \\ &=\sum_{n\geq 2}\pi(n)\left(\log\left(1-\tfrac{1}{n^s}\right)-\log\left(1-\tfrac{1}{(n+1)^s}\right)\right)\\&:=\sum_{n\geq 2}b_n\end{align} Rearrangement is allowed because $\log(1-n^{-s})\sim n^{-\sigma}$ and $\pi(n)\leq n $

$\textbf{Q:}$ Why is this obvious? I considered $$|a_n-b_n|=\left|-\pi(n-1)\log\left(1-\tfrac{1}{n^s}\right)+\pi(n)\log\left(1-\tfrac{1}{(n+1)^s}\right)\right|.$$ Add and subtract term $\pi(n-1)\log(1-\frac{1}{(n+1)^s})$ and use triangle inequality to see $$|a_n-b_n|\leq n^{-\sigma}+\left|\pi(n-1)\left(\log\left(1-\tfrac{1}{(1+n)^s}\right)-\log\left(1-\tfrac{1}{n^s}\right)\right)\right|.$$ The second term is bounded by order $n^{-2\sigma}$. Hence I have tail controlled. However this does not directly yield equality but says tail can be arbitrarily controlled.

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The point is that $$ \left| \sum_{n = 2}^N b_n - \sum_{n = 2}^N a_n \right| = \left| \pi (N) \log \left(1 - \tfrac{1}{(N + 1)^s} \right) \right| \leq CN^{1 - \sigma}$$ for some constant $C > 0$, and for sufficient large $N$.

Since $1 - \sigma < 0$, we have $$ \lim_{N \to \infty} \left| \sum_{n = 2}^N b_n - \sum_{n = 2}^N a_n \right| = 0.$$

So since $\sum_{n = 2}^\infty a_n$ converges, it must be true that $\sum_{n = 2}^\infty b_n$ converges too, and $ \sum_{n = 2}^\infty b_n = \sum_{n = 2}^\infty a_n$.

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  • $\begingroup$ Maybe this is a dumb question. Was this arrangement obvious at first sight? It is clear that it is natural to manipulate sum in that manner though it might not converge. $\endgroup$ – user45765 Dec 27 '18 at 1:20
  • $\begingroup$ @user45765 I had to stare at it for a bit... $\endgroup$ – Kenny Wong Dec 27 '18 at 1:21
  • $\begingroup$ I see. Thanks alot. $\endgroup$ – user45765 Dec 27 '18 at 1:22

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