I'd like to know the estimate of the following sum $$\sum_{n\leq x}\sum_{d|n}\Lambda(d)\frac{\phi(d)}{d} $$ where $\Lambda(d)$ is the von mangoldt function and $\phi(d)$ is the Euler totient function. Do you know how to compute this?
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4$\begingroup$ Note that $\Lambda(d)$ restricts $d = p^k$ for some prime $p$ and positive integer $k$, and $\phi(p^k)/p^k = 1 - \frac{1}{p}$. This should give you a good start on evaluating the sum. Intuitively, you should think of $\phi(n)/n$ as 1 on overage, so your sum is essentially $\sum_{n\leq x} \sum_{d\mid n} \Lambda(d) = \sum_{n\leq x} \log n = x\log x + O(x)$. $\endgroup$– Joshua StuckyCommented Mar 19, 2023 at 5:16
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$\begingroup$ Please use a high-level tag like "nt.number-theory". I added this tag now. $\endgroup$– GH from MOCommented Mar 19, 2023 at 5:37
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Following Joshua Stucky's remark, the sum can be rewritten as the following sum over prime numbers: $$\sum_{p\leq x}(\log p)\left(1-\frac{1}{p}\right)\sum_{k=1}^\infty\left\lfloor\frac{x}{p^k}\right\rfloor.$$ Hence the sum is upper bounded by $$\sum_{p\leq x}(\log p)\left(1-\frac{1}{p}\right)\frac{x}{p-1}=x\sum_{p\leq x}\frac{\log p}{p}=x\log x+O(x),$$ and it is lower bounded by $$\sum_{p\leq x}(\log p)\left(1-\frac{1}{p}\right)\left(\frac{x}{p}-1\right)>x\sum_{p\leq x}\frac{\log p}{p}-x\sum_{p\leq x}\frac{\log p}{p^2}-\sum_{p\leq x}\log p=x\log x+O(x).$$ Therefore, the sum is $$\sum_{n\leq x}\sum_{d|n}\Lambda(d)\frac{\phi(d)}{d}=x\log x+O(x).$$
answered Mar 19, 2023 at 5:36