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I'm trying to understand a proof in Chandrasekharan's Introduction to Analytic Number Theory. Specifically, the proof of the lemma on p.118 before Dirichlet's theorem on primes in arithmetic progressions.

Define

$$ Q(s) = \log P(s) $$

for some particular branch of the logarithm for $\sigma > 1$. If

$$ Q(s) = \sum_{n=1}^{\infty} \frac{a_n}{n^s}, $$

which converges absolutely for $\sigma > 1$ and such that the coefficients $a_n$ are nonnegative, how can I conclude that

$$ P(s) = e^{Q(s)} = 1 + Q(s) + \frac{Q^2(s)}{2!} + \cdots $$

can be written as a Dirichlet series which is convergent for $\sigma > 1$ and whose coefficients are nonnegative?

I know that a product of Dirichlet series with nonnegative coefficients is again a Dirichlet series of nonnegative coefficients which converges on the intersection of the two half-planes of convergence, and hence each of the terms here is a Dirichlet series, but I don't see why an infinite sum of Dirichlet series is necessarily itself a Dirichlet series.

Also, how do I show that, if the Dirichlet series of $Q(s)$ converges, so does the Dirichlet series of $P(s)$, and vice versa?

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1 Answer 1

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The result evidently follows only from absolute convergence of $Q(s)$. As I mentioned in the question, any product of two absolutely convergent Dirichlet series is an absolutely convergent Dirichlet series, so let

$$ \frac{Q^k(s)}{k!} = \sum_{n=1}^{\infty} \frac{a_{k,n}}{n^s}. $$

Then

$$ e^{Q(s)} = \sum_{k=0}^{\infty} \frac{Q^k(s)}{k!} = \sum_{k=0}^{\infty} \sum_{n=1}^{\infty} \frac{a_{k,n}}{n^s}. $$

If we assume $Q(s)$ converges absolutely, then the inner sum converges asbolutely (after which the outer sum clearly does too), so we may swap the order of summation to get

$$ e^{Q(s)} = \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{a_{k,n}}{n^s} = \sum_{n=1}^{\infty} \frac{b_n}{n^s}, $$

where

$$ b_n = \sum_{k=0}^{\infty} a_{k,n}. $$

Conversely, if $\sum_n \frac{b_n}{n^s}$ converges absolutely, then, since the $a_{k,n}$ are all nonnegative, the double sum $\sum_n \sum_k \frac{a_{k,n}}{n^s}$ converges absolutely, and hence switching the order of summation allows us to conclude that each $\sum_n \frac{a_{k,n}}{n^s}$ converges absolutely. In particular, $Q(s)$ converges absolutely, as desired.

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