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I'm stuck on the following problem.

Use the fact that $$\sum_{\substack{p\le x \\ p\,\text{prime}}}\frac{\log p}{p}=\log (x)+O(1)$$ to show that $$\sum_{\substack{n\le x \\ n\in\mathbb{N}}}\frac{\Lambda(n)}{n}=\log (x)+O(1),$$ where $\Lambda$ is the von Mangoldt function.

It looks like it should be easy. Please help; hints are preferred.

My main set back is figuring out what to do with the powers of primes. Here's what I've got so far.

We have $$O(1)=\sum_{\substack{p\le x \\ p\,\text{prime}}}\frac{\log (p)}{p}-\log (x)$$ so there exists a $\lambda>0$ s.t. for any $\varepsilon>0$ there exists $M_{\varepsilon}>0$ s.t. if $x>M_{\varepsilon}$ then $$\left\lvert\sum_{\substack{p\le x \\ p\,\text{prime}}}\frac{\log (p)}{p}-\log (x)\right\rvert\le\lambda\lvert 1\rvert =\lambda.$$

So let $\varepsilon>0$. Now, by definition of $\Lambda$, we have (for $x>M_{\varepsilon}$) that, with $P(x)$ as "$p^k\le x, p\, \text{prime}, k\in\mathbb{N}\backslash\{1\}$,"

$$ \begin{align} \left\lvert\sum_{\substack{p^k\le x \\ p\,\text{prime} \\ k\in\mathbb{N}}} \frac{\log (p)}{p^k}-\log (x)\right\rvert &\le \left\lvert\sum_{\substack{p\le x \\ p\,\text{prime}}} \frac{\log (p)}{p}-\log (x)\right\rvert +\left\lvert\sum_{P(x)}\frac{\log (p)}{p^k}\right\rvert. \\ &\le \lambda + \left\lvert\sum_{P(x)}\frac{\log (p)}{p^k}\right\rvert \\ &= \lambda + \sum_{\substack{p^2\le x \\ p\,\text{prime}}} \frac{\log p}{p^2}+\sum_{\substack{p^3\le x \\ p\,\text{prime}}} \frac{\log p}{p^3}+\dots \\ &= \lambda + \sum_{k=2}^\infty\underbrace{\sum_{\substack{p^k\le x \\ p\,\text{prime}}} \frac{\log p}{p^k}}_{\text{Each of these is less than }\lambda.}\tag{1} \\ &\le\sum_{k\in\mathbb{N}}\lambda. \end{align} $$

But there are infinitely many terms in that series of $\lambda$s, so that bound is too crude (of course). The terms in $(1)$ are each bounded by some (minimal) constant $\sigma_{k, x}$ though so I just need to show that $$\sum_{k=2}^\infty\sigma_{k, x}<\infty$$ if at all possible :/

My aim is to bound it all by a constant.

I've tried other things too but - believe it or not - this seems like my best shot.

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This was the closest thing I could find:… – Shaun Feb 10 '14 at 20:33
I spent time on this crap. It's an instance of when one hopes asking the question helps one answer it. – Shaun Feb 10 '14 at 20:40
Have you tried applying Abel's identity? For an example (and a reference to Abel's identity) see here: – Aryabhata Feb 10 '14 at 21:08
Just pointing out that you introduced a quantity $\epsilon$ that doesn't get used. You can just directly say: there exists $\lambda$ and $M$ such that for $x>M$, your second displayed equation holds. – Greg Martin Feb 11 '14 at 7:07
Yep, there's never a need to introduce a quantity you never use. (Math is about verifying truths, not placating an epsilon overlord!) – Greg Martin Feb 11 '14 at 18:54
up vote 4 down vote accepted

In the written-out term you can bracket the $\log{p}$'s and you get $$S=\sum\limits_{n\leq x}\frac{\Lambda(n)}{n}=\sum\limits_{p\leq x}\log{p}\left(\sum\limits_{p^i\leq x}\frac{1}{p^i}\right)$$ The term in the bracket is bounded by $$\frac{1}{1-\frac{1}{p}}-1=\frac{1}{p-1}$$ So $$S\leq\sum\limits_{p\leq x}\frac{\log p}{p-1}=\sum\limits_{p\leq x}\frac{\log p}{p}+\sum\limits_{p\leq x}\frac{\log p}{p^2-p}=\log x + O(1)$$ since the last sum converges.

Furthermore, because we add terms to it if using the von Mangoldt function instead of the logs of primes, we obtain $$\log x+O(1)=\sum\limits_{p\leq x}\frac{\log p}{p}\leq S$$ and we see that $S=\log x + O(1)$.

I'm not sure if the last step is correct, because I'm not used to the Big O notation, so please reply if this is invalid.

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The last is a little abuse-of-notationish, but that's common with big-Oh notation. You could write $$0 \leqslant S - \sum_{p\leqslant x} \frac{\log p}{p} \leqslant C := \sum_{p} \frac{\log p}{p(p-1)} < \infty.$$ – Daniel Fischer Feb 10 '14 at 22:20
Thank you. I wasn't sure about that bound and the convergence of the second sum at first but I think it checks out. And thank you, @DanielFischer, for the clarification :) – Shaun Feb 11 '14 at 16:41

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