Infinite sum of asymptotic expansions I have a question about an infinite sum of asymptotic expansions:
Assume that $f_k(x)\sim a_{0k}+\dfrac{a_{1k}}{x}+\dfrac{a_{2k}}{x^2}+\cdots$
with $a_{0k}\leq \dfrac{1}{k^2}$, $a_{1k}\leq \dfrac{1}{k^3}$, $a_{2k}\leq \dfrac{1}{k^4}$, $\cdots$.
Does it follow that $\sum_{k=1}^{\infty}f_k(x)\sim \sum_{k=1}^{\infty}a_{0k}+\dfrac{\sum_{k=1}^{\infty}a_{1k}}{x}+\dfrac{\sum_{k=1}^{\infty}a_{2k}}{x^2}+\cdots$?
By the way, could you please suggest me materials to learn asymptotic expansions like this? Thanks
 A: There are a lot of implicit limits you're juggling at once, so I would suspect that this is not true in general.  It does hold though if you have some additional uniformity assumptions.  For instance...
Suppose in addition that


*

*you can find a positive sequence $(b_{kn})$ such that, for all integers $n > 0$,
$$
\sum_{k=1}^{\infty} b_{kn} < \infty,
$$

*and, for integers all $n > 0$, you can find a constant $X_n$ which does not depend on $k$ such that, for every real $x > X_n$ and every integer $k > 0$,
$$
\left| f_k(x) - \sum_{j=0}^{n-1} \frac{a_{jk}}{x^j} \right| \leq \frac{b_{kn}}{x^n}.
$$
If these conditions are satisfied then the result holds.  Indeed, fix $n > 0$ and suppose $x > X_n$.  We have
$$
\begin{align}
\sum_{k=1}^{\infty} f_k(x) &= \sum_{k=1}^{\infty} \left[\sum_{j=0}^{n-1} \frac{a_{jk}}{x^j} + \left( f_k(x) - \sum_{j=0}^{n-1} \frac{a_{jk}}{x^j} \right) \right] \\
&= \sum_{j=0}^{n-1} \frac{1}{x^j} \sum_{k=1}^{\infty} a_{jk} + \sum_{k=1}^{\infty} \left( f_k(x) - \sum_{j=0}^{n-1} \frac{a_{jk}}{x^j} \right).
\end{align}
$$
Each of the sums over the $a_{jk}$ converge by the assumption that $|a_{jk}| \leq k^{-j-2}$.  Further, the last sum is bounded by
$$
\begin{align}
\left| \sum_{k=1}^{\infty} \left( f_k(x) - \sum_{j=0}^{n-1} \frac{a_{jk}}{x^j} \right) \right| &\leq \sum_{k=1}^{\infty} \left|f_k(x) - \sum_{j=0}^{n-1} \frac{a_{jk}}{x^j} \right| \\
&\leq \frac{1}{x^n} \sum_{k=1}^{\infty} b_{kn}.
\end{align}
$$
This implies that
$$
\sum_{k=1}^{\infty} \left( f_k(x) - \sum_{j=0}^{n-1} \frac{a_{jk}}{x^j} \right) = O\!\left(\frac{1}{x^n}\right)
$$
as $x \to \infty$, so we can write
$$
\sum_{k=1}^{\infty} f_k(x) = \sum_{j=0}^{n-1} \frac{1}{x^j} \sum_{k=1}^{\infty} a_{jk} + O\!\left(\frac{1}{x^n}\right)
$$
as $x \to \infty$.  As $n$ was arbitrary, this implies that
$$
\sum_{k=1}^{\infty} f_k(x) \sim \sum_{j=0}^{\infty} \frac{1}{x^j} \sum_{k=1}^{\infty} a_{jk}
$$
as $x \to \infty$.
