Reference for asymptotics on sum

Question

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.

Answer 1

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, $$ \begin{align} \sum_{k=1}^m\binom{m}{k}\frac{a^k}k &\sim\frac{(a+1)^m}m\frac{a+1}a\\ &=\frac{(a+1)^{m+1}}{am}\tag{6} \end{align} $$

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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, $$ \sum_{k=1}^m\binom{m}{k}\frac{a^k}k\sim-\log(m)-\gamma-\log(-a)\tag{9} $$