Big $\mathcal{O}$ Notation inquiry while approximating $\sum \frac{\log n}{n}$

I have a function $S(x)$ which is bounded over and also listed below as adheres to :

$f(x) + C_1 + \mathcal{O}(g(x)) < S(x) < f(x) + C_2 + \mathcal{O}(g(x))$ as $x \rightarrow \infty$

Can I end that $$S(x) = f(x) + C + \mathcal{O}(g(x))$$ as $x \rightarrow \infty$? Can any person offer a brief evidence for this reality?

EDIT :

(To clear the complication, I am mentioning the trouble listed below. I am asking yourself if what I did was right.)

Confirm that as $x \rightarrow \infty$

$$\displaystyle \sum_{n \leq x} \frac{\log n}{n} = \frac{(\log x)^2}{2} + C + \mathcal{O}(\frac{\log(x)}{x})$$

where $C$ is a constant.

This is just how I did it.

Approximate the summation as an indispensable $\int \frac{\log(x)}{x} dx$ from above and also from below. (The common means of thinking of bounds for $p$ collection). After that we will certainly get the lower bound as $\frac{\log([x])^2}{2} + C_1$ and also the upper bound as $\frac{\log([x])^2}{2} + C_2$.

After that, $\log([x]) = \log(x) + \log(\frac{[x]}{x}) = \log(x) + \log(1-\frac{\{x\}}{x}) = \log(x) + \mathcal{O}(\frac{\{x\}}{x}) = \log(x) + \mathcal{O}(\frac{1}{x})$.

Connecting this right into the above I get the the inquiry I asked.

I did this when I was asked on a timed test given that I can not assume of various other means to reach the solution. Various other solutions and also pointers invite as constantly.

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2019-12-02 02:51:02
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Answers: 4

I do not assume so due to the fact that $g(x)$ could be also tiny. Take $f(x)=0, C_1=-1.5, C_2=1.5, g(x)=\exp(-x)$. After that does not $S(x)=\sin(x)$ breach it?

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2019-12-03 04:24:31
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If $a<x<b$, after that $|x|\le \max \lbrace |a|,|b|\rbrace$.

Consequently, if $C_1=C_2=C$, what you have there indicates

$$ |S(x)-f(x)-C|\le K |g(x)| $$

for some $K>0$, i.e. $S(x)=f(x)+C+O(g(x))$.

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2019-12-03 04:24:11
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A false impression requires to be cleared below. Remember that $f(x) = O(g(x))$ suggests that there exists a constant $C$ such that $|f(x)| \le C |g(x)|$ in an area of whatever value of $x$ you want. It is a misuse of symbols that we make use of the equal rights icon below in all ; $O(g(x))$ is not a function, so it can not be "equal" to $f(x)$ in the common feeling. See the conversation at the Wikipedia article.

Based upon this definition, individuals occasionally claim that $f(x) = h(x) + O(g(x))$ if $f(x) - h(x) = O(g(x))$. This is an added misuse of symbols, yet it usually does not create troubles and also can be exceptionally hassle-free. I presume you can write $f(x) < h(x) + O(g(x))$ if you actually intended to, yet this is repetitive, given that the inequality is currently constructed right into the definition of large - O symbols.

Yet I have no suggestion what $f(x) > h(x) + O(g(x))$ is intended to suggest. If you are attempting to claim that there is a constant $C$ such that $|f(x) - h(x)| > C |g(x)|$, the proper means to write this is making use of large - Omega, not large - O, as $f(x) = h(x) + \Omega(g(x))$.

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2019-12-03 04:23:37
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Since you desired a various evidence strategy, you can attempt making use of Abel's Identity, which has actually become fairly valuable in analytic number theory.

As an example see this : An estimate for sum of reciprocals of square roots of prime numbers.

To use this to your inquiry :

Since we understand that

$\displaystyle \sum_{1 \le n \le x} \frac1{n} = \log x + \gamma + R(x)$, where $\displaystyle R(x) = \mathcal{O}\left(\frac1{x}\right)$

Using Abel is identification we have that

$\displaystyle \sum_{1 \le n \le x} \frac{\log n}{n} = (\log x+ \gamma + R(x))\log x - \int_{1}^{x} \frac{\log t+ \gamma + R(t)}{t} \ \text dt$

i.e.

$\displaystyle \sum_{1 \le n \le x} \frac{\log n}{n} = \frac{\log^2 x}{2} + R(x)\log x - \int_{1}^{x} \frac{R(t)}{t} \ \text dt$

Since $\displaystyle R(x) = \mathcal{O}\left(\frac1{x}\right)$ we have that $\displaystyle \int_{1}^{\infty} \frac{R(t)}{t} \ \text dt = \eta$ exists. We additionally have that $\displaystyle R(x) \log x \to 0 \ \text{as} \ x \to \infty$.

Hence we have that

$\displaystyle \sum_{1 \le n \le x} \frac{\log n}{n} = \frac{\log^2 x}{2} -\eta + \mathcal{O}\left(\frac{\log x}{x}\right) \ \ \text{as} \ \ x \to \infty$

Another valuable strategy is to attempt making use of the Euler-Maclaurin Summation formula.

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2019-12-03 04:12:47
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