Exponentiation of a Dirichlet series 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?
 A: 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.
