What does the convergence of a Dirichlet series tells us about the convergence of a power series? If $D(s)=\displaystyle \sum_{k\geqslant 1} f(k)\, k^{s}$ converges for $\Re(s)\lt a$, what is the radius of convergence of $\displaystyle \sum_{k\geqslant 1}f(k)\, x^k$ $=T(x)$? 
Conversely, what does the radius of convergence of $T(x)$ tells us about the plane $D(s)$ converges?
 A: There is no strong connection between the radius of convergence of a power series and the abscissa of convergence (or of absolute convergence) of the Dirichlet series with the same coefficients.
Let
$$R = \sup \Biggl\{ \lvert z\rvert : \sum_{k = 1}^{\infty} f(k) z^k \text{ converges}\Biggr\}$$
and
$$\sigma = \sup \Biggl\{ \operatorname{Re} s : \sum_{k = 1}^{\infty} f(k)k^s \text{ converges}\Biggr\}.$$
Then we have the following implications:


*

*$R < 1 \implies \sigma = -\infty$,

*$R > 1 \implies \sigma = +\infty$,

*$\sigma > -\infty \implies R \geqslant 1$,

*$\sigma < +\infty \implies R \leqslant 1$,


and what follows from them (e.g. $-\infty < \sigma < +\infty \implies R = 1$). When $R = 1$, that doesn't tell us anything about $\sigma$.
The point is that $\lvert z\rvert^k$ grows resp. decays exponentially for $\lvert z\rvert \neq 1$, and to overcome that to have $R \neq 1$ one must have exponential decay resp. growth [growth only for a subsequence] of the coefficients. On the other hand, $k^{\operatorname{Re} s}$ grows resp. decays much slower [polynomially], and thus can only overcome at most polynomial growth/decay of the coefficients.
Making these observations formal:
Suppose $R < 1$. By the Cauchy-Hadamard formula, for every $c \in (1, R^{-1})$ there is a sequence $(k_m)$ of indices such that $\lvert f(k_m)\rvert \geqslant c^{k_m}$ for all $m$. But that implies $\lvert f(k_m)k_m^s\rvert \xrightarrow{m\to\infty} +\infty$ for all $s\in \mathbb{C}$, so the Dirichlet series cannot converge anywhere.
Suppose $R > 1$. By Cauchy-Hadamard, for every $c \in (R^{-1},1)$ there is a $K \in [0,+\infty)$ such that $\lvert f(k)\rvert \leqslant K\cdot c^k$ for all $k$. Then $f(k)k^s \to 0$ for all $s\in \mathbb{C}$, and hence the Dirichlet series converges on all of $\mathbb{C}$.
Suppose $\sigma > -\infty$. Then $f(k)k^t \to 0$ for every $t < \sigma$, in particular this sequence is bounded, so there is a $B\in (0,+\infty)$ such that $\lvert f(k)\rvert \leqslant B\cdot k^{-t}$ for all $k$. Then
$$\limsup_{k\to\infty} \lvert f(k)\rvert^{1/k} \leqslant \lim_{k\to \infty} B^{1/k}k^{-t/k} = 1,$$
so $R \geqslant 1$.
Suppose $\sigma < +\infty$. Then $f(k)k^{t}$ is unbounded for all $t > \sigma + 1$ (otherwise the Dirichlet series would converge absolutely for $\operatorname{Re} s < t - 1$), and we find a subsequence with $\lvert f(k_m)\rvert \geqslant k_m^{-t}$ for all $m$. Then
$$\limsup_{k \to \infty} \lvert f(k)\rvert^{1/k} \geqslant \lim_{m\to \infty} k_m^{-t/k_m} = 1,$$
i.e. $R \leqslant 1$.
Choosing $f(k) = c^{\sqrt{k}}$ we have $R = 1$ when $c\neq 0$, and $\sigma = -\infty$ for $\lvert c\rvert > 1$ while $\sigma = +\infty$ for $0 < \lvert c\rvert < 1$. Everything in between is of course also possible for $R = 1$.
