show that $\sum_{p\leq x} \frac{1}{p \log p} = O(1)$

Question

Please my knowledge in this field is very low so could you help me solve this question in analytic number theory.

Answer 1

Use partial summation, or Abel's summation formula, which states that if $f$ is continuous and \[A(x) = \sum_{n \leq x} a_n,\] then \[\sum_{n \leq x} a_n f(n) = f(x) A(x) - \int_{1}^{x} f'(t) A(t) \, dt.\] Now take \[f(x) = \frac{1}{x \log x},\] so that \[f'(x) = - \frac{\log x + 1}{x^2 (\log x)^2},\] and \[a_n = \begin{cases} 1 & \text{if $n = p$ is a prime,} \\\ 0 & \text{otherwise,} \end{cases}\] and use the fact that \[\pi(x) = \sum_{p \leq x} 1 = O\left(\frac{x}{\log x}\right).\]

Of course, other choices of $f$ and $a_n$ would also work.

Here's some more detail. Using our choice of $f$ and $a_n$, so that $A(x) = \pi(x)$, we get \[\sum_{p \leq x} \frac{1}{p \log p} = \frac{1}{x \log x} \pi(x) + \int_{2}^{x} \frac{\log t + 1}{t^2 (\log t)^2} \pi(t) \, dt.\] Now as $\pi(x) = O(x/\log x)$, we can just plug this in to get \[\sum_{p \leq x} \frac{1}{p \log p} = O\left(\frac{1}{x \log x} \frac{x}{\log x}\right) + O\left(\int_{2}^{x} \frac{\log t + 1}{t^2 (\log t)^2} \frac{t}{\log t} \, dt\right).\] The first term is \[O\left(\frac{1}{(\log x)^2}\right),\] while the second is \[O\left(\int_{2}^{x} \frac{1}{t (\log t)^2} \, dt + \int_{2}^{x} \frac{1}{t(\log t)^3} \, dt\right),\] and by making the substitution $u = \log t$, one finds that \[\int_{2}^{x} \frac{1}{t (\log t)^2} \, dt = \frac{1}{\log x} - \frac{1}{\log 2},\] \[\int_{2}^{x} \frac{1}{t (\log t)^3} \, dt = \frac{2}{(\log x)^2} - \frac{2}{(\log 2)^2}.\] As $1/\log 2$ is a constant, this shows that \[\sum_{p \leq x} \frac{1}{p \log p} = O(1).\]

Answer 2

If we know that $\pi(x+y)-\pi(x)\leq\frac{2y}{\log y}$ then it follows that:

$$ \sum_{p\leq x}\frac{1}{p \log p}\leq \sum_{\alpha=1}^{\left\lceil \log_2 x\right\rceil}\sum_{p\in[2^\alpha,2^{\alpha+1})}\frac{1}{p\log p}\leq \sum_{\alpha=1}^{\left\lceil \log_2 x\right\rceil}\frac{2^{\alpha+1}}{\alpha\log 2\cdot\left( 2^{\alpha}\log 2^{\alpha}\right)} $$ hence: $$ \sum_{p\leq x}\frac{1}{p \log p}\leq \frac{2}{\log^2 2}\sum_{\alpha=1}^{\left\lceil \log_2 x\right\rceil}\frac{1}{\alpha^2}\leq\frac{\pi^2}{3\log^2 2}.$$

Answer 3

We want to show that \begin{align*} \sum_{p\leq x}\frac{1}{p\log p}=\mathrm{O}(1). \end{align*} Using Abel's Summation formula, let $a_{n}=\frac{1}{n}$ and $f(t)=\frac{1}{\log t}$. Then we have \begin{align*} A(x)=\sum_{n\leq x}a_{n}=\sum_{n\leq x}\frac{1}{n}=\log\log x + \alpha +\mathrm{O}\left(\frac{1}{\log x}\right) \end{align*} and \begin{align*} f'(t)= -\frac{1}{t\left(\log t\right)^2}. \end{align*} Then, \begin{align*} \sum_{p\leq x}\frac{1}{p\log p}&=\sum_{n\leq x}a_{n}f(n)=A(x)f(x)-\int_{1}^{x}A(t)f'(t)\; dt \\ &= \frac{\log \log x +\alpha +\mathrm{O}\left(\frac{1}{\log x}\right)}{\log x}+\int_{1}^{x}\left(\frac{\log \log x +\alpha +\mathrm{O}\left(\frac{1}{\log x}\right)}{t\left(\log x\right)^2}\right)\; dt\\ &= \frac{\log \log x}{\log x}+\frac{\alpha}{\log x}+\mathrm{O}\left(\frac{1}{(\log x)^2}\right)+\int_{1}^{x}\frac{\log \log t}{t\left(\log t\right)^2}\; dt +\alpha\int_{1}^{x}\frac{1}{t\left(\log t\right)^2}\; dt +\int_{1}^{x}\frac{\mathrm{O}\left(1/\log t\right)}{t\left(\log t\right)^2}\; dt \\ &= \frac{\log \log x}{\log x}+\frac{\alpha}{\log x}+\mathrm{O}\left(\frac{1}{(\log x)^2}\right)+\int_{2}^{x}\frac{\log \log t}{t\left(\log t\right)^2}\; dt +\alpha\int_{2}^{x}\frac{1}{t\left(\log t\right)^2}\; dt +\int_{2}^{x}\frac{\mathrm{O}\left(1/\log t\right)}{t\left(\log t\right)^2}\; dt \\ &= \frac{\log \log x}{\log x}+\frac{\alpha}{\log x}+\mathrm{O}\left(\frac{1}{(\log x)^2}\right)+\int_{2}^{x}\frac{\log \log t}{t\left(\log t\right)^2}\; dt +\alpha\int_{2}^{x}\frac{1}{t\left(\log t\right)^2}\; dt +\mathrm{O}\left(\int_{2}^{x}\frac{1}{t\left(\log t\right)^3}\; dt\right)\\ &= \frac{\log \log x}{\log x}+\frac{\alpha}{\log x}+\mathrm{O}\left(\frac{1}{(\log x)^2}\right)+\int_{2}^{x}\frac{\log \log t}{t\left(\log t\right)^2}\; dt +\alpha\int_{2}^{x}\frac{1}{t\left(\log t\right)^2}\; dt +\mathrm{O}\left(\frac{1}{\left(\log t\right)^2}\right)\\ &= \frac{\log \log x}{\log x}+\frac{\alpha}{\log x}+\int_{2}^{x}\frac{\log \log t}{t\left(\log t\right)^2}\; dt +\alpha\int_{2}^{x}\frac{1}{t\left(\log t\right)^2}\; dt +\mathrm{O}\left(\frac{1}{\left(\log t\right)^2}\right) \end{align*} But \begin{align*} \int_{2}^{x}\frac{\log \log t}{t\left(\log t\right)^2}\; dt = \left(-\frac{\log \log t}{\log t}-\frac{1}{\log t}\right)\Big|_{2}^{x} = -\frac{\log \log x}{\log x}+ \frac{\log \log 2}{\log 2} -\frac{1}{\log x}+\frac{1}{\log 2}. \end{align*} and \begin{align*} \alpha\int_{2}^{x}\frac{1}{t\left(\log t\right)^2}\; dt &=\alpha\int_{2}^{\infty}\frac{1}{t\left(\log t\right)^2}\; dt -\alpha\int_{x}^{\infty}\frac{1}{t\left(\log t\right)^2}\; dt \\ &= -\frac{\alpha}{\log t}\Big|_{2}^{\infty}+\frac{\alpha}{\log t}\Big|_{x}^{\infty}\\ &= \frac{\alpha}{\log 2} -\frac{\alpha}{\log x}. \end{align*} Hence \begin{align*} \sum_{p\leq x}\frac{1}{p\log p}&=\frac{\log \log x}{\log x}+\frac{\alpha}{\log x}-\frac{\log \log x}{\log x}+ \frac{\log \log 2}{\log 2} -\frac{1}{\log x}+\frac{1}{\log 2}+\frac{\alpha}{\log 2} -\frac{\alpha}{\log x} +\mathrm{O}\left(\frac{1}{\left(\log t\right)^2}\right)\\ &= \frac{\log \log 2}{\log 2} +\frac{1}{\log 2}+\frac{\alpha}{\log 2} -\frac{1}{\log x}+\mathrm{O}\left(\frac{1}{\left(\log t\right)^2}\right)\\ &= \mathrm{O}(1),\; \text{as}\; x\rightarrow \infty . \end{align*}