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Consider the following equality $$ \sum_{n=1}^{\infty} \frac{\Lambda(n)}{\log n} n^{-s} = -\log(s-1)+ \sum_{\rho} \log(s- \rho) + \dots$$

Why is an approximation of $\Lambda(n)$ the following: $$\Lambda(n) \approx 1- \sum_{\rho} n^{\rho-1}$$

Using the fact that $$\log (s- \rho) \approx -\sum_{n=1}^{\infty} \frac{n^{\rho-s-1}}{\log n}$$

we have $$ \sum_{n=1}^{\infty} \frac{\Lambda(n)}{\log n} n^{-s} \approx \sum_{n=1}^{\infty} \frac{n^{-s}}{\log n} - \sum_{\rho} \sum_{n=1}^{\infty} \frac{n^{\rho-s-1}}{\log n}$$

From here, how do we solve for $\Lambda(n)$?

Added. I did the following $$ \sum_{n=1}^{\infty} \frac{\Lambda(n)}{\log n} n^{-s} \approx \sum_{n=1}^{\infty} \frac{n^{-s}}{\log n} - \sum_{n=1}^{\infty} \sum_{\rho} \frac{n^{\rho-s-1}}{\log n}$$

$$ \approx \sum_{n=1}^{\infty} \left(\frac{n^{-s}}{\log n} - \sum_{\rho} \frac{n^{\rho-s-1}}{\log n} \right)$$

i.e. interchanged sums so that each double sum begins with index $n$.

Then $$\Lambda(n) \approx n^{s} \log n \cdot \left(\frac{n^{-s}}{\log n} - \sum_{\rho} \frac{n^{\rho-s-1}}{\log n} \right)$$

Added I solved it.

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I did the following $$ \sum_{n=1}^{\infty} \frac{\Lambda(n)}{\log n} n^{-s} \approx \sum_{n=1}^{\infty} \frac{n^{-s}}{\log n} - \sum_{n=1}^{\infty} \sum_{\rho} \frac{n^{\rho-s-1}}{\log n}$$

$$ \approx \sum_{n=1}^{\infty} \left(\frac{n^{-s}}{\log n} - \sum_{\rho} \frac{n^{\rho-s-1}}{\log n} \right)$$

i.e. interchanged sum sums so that each double sum begins with index $n$.

Then $$\Lambda(n) \approx n^{s} \log n \cdot \left(\frac{n^{-s}}{\log n} - \sum_{\rho} \frac{n^{\rho-s-1}}{\log n} \right)$$

$$ \approx 1- \sum_{\rho} n^{\rho-1}$$

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