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.


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

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?


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

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


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

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

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


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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.