Reference for asymptotics on sum Quite simply I'm looking for the large $m$ asymptotic behavior of
\begin{equation}
\sum_{k=1}^{m}{m\choose k}\frac{a^k}{k}
\end{equation}
where $a$ is a constant. This looks easy for someone who knows about this area, so just looking to be pointed in the right direction.
 A: If $\boldsymbol{a\gt0}$
Note that
$$
\begin{align}
f(a,m)
&=\sum_{k=1}^m\binom{m}{k}\frac{a^k}k\\
&=\sum_{k=1}^m\left[\binom{m-1}{k-1}+\binom{m-1}{k}\right]\frac{a^k}k\\
&=\frac1m\sum_{k=1}^m\binom{m}{k}a^k+\sum_{k=1}^{m-1}\binom{m-1}{k}\frac{a^k}k\\
&=\frac{(a+1)^m-1}m+f(a,m-1)\tag{1}
\end{align}
$$
Therefore,
$$
\sum_{k=1}^m\binom{m}{k}\frac{a^k}k=\sum_{k=1}^m\frac{(a+1)^k-1}k\tag{2}
$$
Then we have
$$
\begin{align}
\lim_{m\to\infty}\frac{m}{(a+1)^m}\sum_{k=1}^m\frac{(a+1)^k}k
&=\lim_{m\to\infty}\frac{m}{(a+1)^m}\sum_{k=0}^{m-1}\frac{(a+1)^{m-k}}{m-k}\\
&=\color{#C00000}{\lim_{m\to\infty}\sum_{k=0}^{m-1}(a+1)^{-k}}+\color{#00A000}{\lim_{m\to\infty}\sum_{k=0}^{m-1}\frac{k(a+1)^{-k}}{m-k}}\\
&=\color{#C00000}{\frac{a+1}a}+\color{#00A000}{0}\tag{3}
\end{align}
$$
and
$$
\begin{align}
\lim_{m\to\infty}\frac{m}{(a+1)^m}\sum_{k=1}^m\frac1k
&\le\lim_{m\to\infty}\frac{m^2}{(a+1)^m}\\
&=0\tag{4}
\end{align}
$$
Therefore,
$$
\lim_{m\to\infty}\frac{m}{(a+1)^m}\sum_{k=1}^m\binom{m}{k}\frac{a^k}k
=\frac{a+1}a\tag{5}
$$
Thus, asymptotically,
$$
\begin{align}
\sum_{k=1}^m\binom{m}{k}\frac{a^k}k
&\sim\frac{(a+1)^m}m\frac{a+1}a\\
&=\frac{(a+1)^{m+1}}{am}\tag{6}
\end{align}
$$

If $\boldsymbol{-2\lt a\lt0}$
Equation $(2)$ still holds, but now
$$
\lim_{m\to\infty}\sum_{k=1}^m\frac{(a+1)^k}k=-\log(-a)\tag{7}
$$
and
$$
\sum_{k=1}^m\frac1k=\log(m)+\gamma+O\left(\frac1m\right)\tag{8}
$$
Therefore, asymptotically,
$$
\sum_{k=1}^m\binom{m}{k}\frac{a^k}k\sim-\log(m)-\gamma-\log(-a)\tag{9}
$$
