Theorem 3.16. in Analytic Number Theory by Apostol

Question

The below texts are from the book Introduction to Analytic Number Theory by Apostol:

I have two questions which I couldn't find solutions for them:

$1-$ According to Thm 3.16., $\sum_{n\le x} \Lambda(n) \Big[ \dfrac{x}{n} \Big] = \sum_{p\le x} \ln(p) \Big[ \dfrac{x}{p} \Big] + \sum_{p\le x} \sum_{m=2}^{\infty} \ln(p) \Big[ \dfrac{x}{p^m} \Big] = \sum_{p\le x} \ln(p) \Big[ \dfrac{x}{p} \Big] + O(x),$ so by Eq. (22) we have $x \ln(x) - x + O(\ln (x)) = \sum_{p\le x} \ln(p) \Big[ \dfrac{x}{p} \Big] + O(x),$ so $x \ln(x) + O(\ln (x)) + O(x) = \sum_{p\le x} \ln(p) \Big[ \dfrac{x}{p} \Big],$ so if the Eq. (23) is true then it implies that the book has considered $O(\ln (x)) + O(x) = O(x)$ but it is not true since $O(\ln (x)) + O(x) = O(\ln (x)).$ Am I wrong?

$2-$ How to prove that the sum $\sum_{n=2}^{\infty} \dfrac{\ln(n)}{n(n-1)}$ is finite?

Answer 1

1. No, it is not true that $O(\ln x) + O(x) = O(\ln x)$. For instance, $x$ is $O(\ln x) + O(x)$, but $x \gg \ln x$ and so $x$ is not $O(\ln x)$.

2. As $\ln x \ll x^{\alpha}$ for any $\alpha > 0$, we have that $$ \sum_{n \geq 2} \frac{\ln n}{n(n-1)} \ll \sum_{n \geq 2} \frac{n^{\alpha}}{n(n-1)} \sim \sum_{n \geq 2} n^{\alpha - 2},$$ for any small $\alpha > 0$, and which converges. Intuitively, logs are so much smaller than polynomials that they barely affect convergence.

Answer 2

(i). For $x\geq 1$ we have $\ln x=\int_1^x (1/y)\;dy\leq \int_1^x (1)\;dy=x-1<x.$

(ii). From (i), we have : If $x\geq 1$ and $k>1$ then $$k x^{1/k}>k(x^{1/k}-1)\geq k\ln x^{1/k}=\ln x.$$ Therefore $0\leq (\ln x)/x<1/x^{k-1}$ and conclude that $\lim_{x\to \infty}(\ln x)/x=0.$ That is, $\ln x=o(x)$ as $x\to \infty.$

$$\text {(iii). }\; n\geq 2 \implies n-1\geq n/2 \implies 1/n(n-1) \leq 2/n^2.$$ From (ii) with $k=2$ we have $0<\ln n< 2 n^{1/2}$ for all sufficiently large $n.$ So for all but finitely many $n\in N$ we have $$0<(\ln n)/n(n-1)<4/n^{3/2}.$$ And the Cauchy Condensation Test shows that $\sum_{n\in N}(1/n^p)<\infty$ for all $p>1.$

Answer 3

You might want to review the definition of big O notation. Remember that $f(x)=O(g(x))$ if and only if $\dfrac{f(x)}{g(x)}$ is bounded for all $x\geqslant a$.

Now, suppose $f_1(x)=O(\ln x),f_2(x)=O(x)$. Then $f_1(x)\leqslant M_1\ln(x),f_2(x)\leqslant M_2x$. Notice that $\ln x\leqslant x$. Then $$O(\ln x)+O(x)=f_1(x)+f_2(x)\leqslant Mx,$$ where $M=\max\{M_1,M_2\}x.$

which, by definition, implies $O(\ln x)+O(x)=O(x)$.