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Some context: I've been studying Chebyshev's $\psi$ - function, which claims that $\psi(x) = \sum_{n \le x} \Lambda(n) = \sum_{p^k \le x} \log p$ where $p$ is prime and $\Lambda(n)$ is the von Mangoldt function.

In my text book, I have that $\sum_{p^k \le x, k \geq 1} \frac{\log p}{p^k} = \sum_{p \le x} \frac{\log p}{p} + \sum_{p^k \le x, k \geq 2} \frac{\log p}{p^k}$.

How and why can the summation be split in this way?

Thanks in advance.

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The left side is a sum over prime powers. The right side splits this into the sum over primes, and the sum over "proper powers", $p^k$ with $p$ prime and $k\ge2$. –  Gerry Myerson Mar 17 at 12:11
    
And the reason to split it like this is that the second term is bounded by a constant (as the infinite series $\sum_{p^k,k\ge2}\frac{\log p}{p^k}$ converges), so one can ignore it when discussing the asymptotic behaviour of the function. –  Emil Jeřábek Mar 20 at 14:31

1 Answer 1

The sum over $k$ on the left is split into the terms with $k=1$ and the terms with $k\gt1$. The terms with $k=1$ give you the first sum on the right, and the terms with $k\gt1$ (thus, $k\ge2$) give you the second sum on the right.

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