Show that $x \left(\sum\limits_{n\leq x} \frac{\ln(n)}{n^2}\right) +\theta(x)=O(x)$

Question

I know that $$x\left(\sum\limits_{n\leq x} \frac{\ln(n)}{n^2}\right)+\theta(x) \leq x\left(\sum\limits_{n\leq x} \frac{\ln(n)}{n}\right)+\theta(x)=x\frac{1}{2}\left(\ln{x}\right)^2+Cx+O(\ln x)+O(x)$$

where $\theta(x)=\sum\limits_{p\leq x} \ln p$

Is that equal to $O(x(\ln x)^2)$ not $O(x)$?

Answer 1

Linking this question purely for the links to the theoretical material (and to some extent, similarity).

From Euler's summation formula

$$\sum\limits_{y< n\leq x}f(n)= \int\limits_{y}^{x}f(t)dt + \int\limits_{y}^{x}\{t\}f'(t)dt-\{x\}f(x)+\{y\}f(y) \tag{1}$$

and

$$\int \frac{\ln{x}}{x^2} dx=-\frac{\ln{x} + 1}{x} + C \tag{2}$$

we have ($n=1$ is trivial, we will consider $n\geq2$) $$\sum\limits_{2\leq n\leq x}\frac{\ln{n}}{n^2}= \int\limits_{2}^{x}\frac{\ln{t}}{t^2}dt+\int\limits_{2}^{x}\{t\}\frac{1-2\ln{t}}{t^3}dt-\{x\}\frac{\ln{x}}{x^2}=\\ \frac{\ln{2}+1}{2}-\frac{\ln{x}+1}{x}+\int\limits_{2}^{x}\{t\}\frac{1-2\ln{t}}{t^3}dt-\{x\}\frac{\ln{x}}{x^2} \tag{3}=...$$

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Now, $\forall x\geq2$ $$\left|\{x\}\frac{\ln{x}}{x^2}\right|\leq \frac{\ln{x}}{x^2}$$ and (it's worth noting that $2\ln{2}-1>0$) $$\left|\int\limits_{2}^{x}\{t\}\frac{1-2\ln{t}}{t^3}dt\right| \leq \int\limits_{2}^{x}\left|\{t\}\frac{1-2\ln{t}}{t^3}\right|dt \leq \int\limits_{2}^{x}\left|\frac{1-2\ln{t}}{t^3}\right|dt=\\ \int\limits_{2}^{x}\frac{2\ln{t}-1}{t^3}dt=-\frac{\ln{x}}{x^2}+\frac{\ln{2}}{2^2}$$ this shows that $\int\limits_{2}^{\infty}\{t\}\frac{1-2\ln{t}}{t^3}dt=I$ exists (by taking $\lim\limits_{x\rightarrow\infty}$) and $$\int\limits_{2}^{x}\{t\}\frac{1-2\ln{t}}{t^3}dt=I-\int\limits_{x}^{\infty}\{t\}\frac{1-2\ln{t}}{t^3}dt= I+\int\limits_{x}^{\infty}\{t\}\frac{2\ln{t}-1}{t^3}dt\leq \\ I+\int\limits_{x}^{\infty}\frac{2\ln{t}-1}{t^3}dt=I+\frac{\ln{x}}{x^2}$$

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Continuing from $(3)$ $$...=\frac{\ln{2}+1}{2}+I-\frac{\ln{x}+1}{x}+O\left(\frac{\ln{x}}{x^2}\right)= E-\frac{\ln{x}+1}{x}+O\left(\frac{\ln{x}}{x^2}\right) \tag{3a}$$ where $E$ is a constant. From $(3a)$ we see that $$\lim\limits_{x\rightarrow\infty} \left(\sum\limits_{2\leq n\leq x}\frac{\ln{n}}{n^2}\right)=E \tag{4}$$ which simply means $$\sum\limits_{2\leq n\leq x}\frac{\ln{n}}{n^2}=O(1) \tag{5}$$

Also, Chebyshev's function has the property that $$\lim\limits_{x\rightarrow\infty}\frac{\theta(x)}{x}=1 \Rightarrow \theta(x)=O(x) \tag{6}$$

All these together $$x\left(\sum\limits_{2\leq n\leq x}\frac{\ln{n}}{n^2}\right)+\theta(x)=xO(1)+O(x)=O(x)$$