Show $\sum_p\frac{1}{p^{1+i}}$ converges

Question

I am struggling with exercise I.15 in Tenenbaum's Analytic Number Theory book.

The problem says to show that (here $\color{blue}{i}$ is the imaginary unit) $$\sum_{p}\frac{1}{p^{1+i}}$$ converges using the strong form of the prime number theorem: $$\pi(x)=\frac{x}{\log(x)}+O\left(\frac{x}{(\log(x))^2}\right).$$ I had no problem showing $\sum_p\frac{1}{p^{1+\epsilon}}$ converges for $\epsilon>0$ using that $p_n\sim n\log(n)$ from the prime number theorem, but this is less obvious.

Answer 1

$$ \sum_{p\leqslant N}p^{-1-i}=\sum_{n=2}^N\big(\pi(n)-\pi(n-1)\big)n^{-1-i} \\=\pi(N)N^{-1-i}+\sum_{n=2}^{N-1}\pi(n)\big(n^{-1-i}-(n+1)^{-1-i}\big) $$ and, since $n^{-1-i}-(n+1)^{-1-i}=(1+i)n^{-2-i}+O(n^{-3})$ as $n\to\infty$, the provided asymptotics for $\pi(n)$ implies that the convergence of $\sum_p p^{-1-i}$ is equivalent to the convergence of $$\sum_{n=2}^\infty\frac1{n^{1+i}\log n}.$$

To see the latter, one might apply the Euler–Maclaurin summation formula $$\sum_{n_1\leqslant n\leqslant n_2}f(n)=\int_{n_1}^{n_2} f(x)\,dx+\frac{f(n_1)+f(n_2)}2+\int_{n_1}^{n_2}\left(\{x\}-\frac12\right)f'(x)\,dx$$ to $f(x)=1/(x^{1+i}\log x)$. Substituting $x=e^t$ in the first integral, we get $$\sum_{2\leqslant n\leqslant N}\frac1{n^{1+i}\log n}=\int_{\log 2}^{\log N}\frac{e^{-it}}{t}\,dt+\frac{f(2)+f(N)}{2}+\int_2^N\left(\{x\}-\frac12\right)f'(x)\,dx.$$

All the terms have a finite limit as $N\to\infty$; e.g. the last one because of $|f'(x)|=O(x^{-2})$.

Of course, the same works with $i$ replaced by $i\tau$ for $\tau\in\mathbb{R}_{\neq0}$. As for computations, we can use $\sum_p p^{-s}=\sum_{n=1}^\infty\big(\mu(n)/n\big)\log\zeta(ns)$ (once we know this converges) or even better, as suggested in section 2.1 here, $$\sum_p p^{-s}=\sum_{p\leqslant A}p^{-s}+\sum_{n=1}^\infty\frac{\mu(n)}n\log\zeta_A(ns),$$ where $\zeta_A(s)=\zeta(s)\prod_{p\leqslant A}(1-p^{-s})$ and $A$ is suitably (moderately) large.

Answer 2

Using the fact that

$$ \int_2^x{\mathrm dt\over\log t}={x\over\log x}+O\left(x\over\log^2x\right), $$

it follows from the PNT in OP's question that for all $x\ge y\ge2$,

$$ \pi(x)-\pi(y)=\sum_{y<p\le x}1=\int_y^x{\mathrm dt\over\log t}+R(y,x), $$

for some $R(y,x)\ll y/\log^2y$, so it follows from partial summation that

$$ \begin{aligned} \sum_{y<p\le x}{1\over p^{1+i}} &=\int_y^x{\mathrm d[\pi(x)-\pi(y)]\over t^{1+i}}=\int_y^x{\mathrm dt\over t^{1+i}\log t}+[t^{-1-i}R(y,t)]_y^x \\ &+(1+i)\int_y^x t^{-2-i}R(y,t)\mathrm dt=A_1+A_2+A_3. \end{aligned} $$

Clearly $A_2+A_3\ll(\log y)^{-2}$, and it follows from the integration by parts that

$$ A_1=\left.{it^{-i}\over\log t}\right|_y^x+i\int_y^x{\mathrm dt\over t^{1+i}\log^2t}\ll{1\over\log y}, $$

so we have

$$ \sum_{y<p\le x}{1\over p^{1+i}}\ll{1\over\log y}, $$

Therefore, it follows from Cauchy's criterion that the series $\sum_pp^{-1-i}$ converges.