As above, $\langle x\rangle$ denotes the fractional part of $x$.

The behavior of $f_n(t):=\sum_{k=0}^\infty\mu^k(1-\mu^k t)^n$ as $n\to\infty$ in the interval $(0,1)$ is as follows. For a subsequence $n_k$ of the natural numbers, the sequence $n_kf_{n_k}(t)$ converges if and only if $\langle-\log_\mu(n_k-1)\rangle$ converges as $k\to\infty$. Moreover, if $\langle-\log_\mu(n_k-1)\rangle\to q\in [0,1]$ as $k\to\infty$, then $\lim_{k\to\infty}n_kf_{n_k}(t)=F_q(t)$ uniformly on compacts of $(0,1)$, where
$$
F_q(t)=\frac{\mu t}{1-\mu}\int_0^\infty\mu^{-\langle \log_\mu(s)+q\rangle}se^{-st}ds
=\sum_{k=-\infty}^\infty\mu^{k-q}e^{t\mu^{k-q}}.
$$

Since $\langle-\log_\mu n\rangle_{n=1}^\infty$ is dense in $[0,1]$, it follows that all the limit points of the sequence $n f_n(t)$ are precisely the functions $F_q(t)$, $0\leq q<1$. The same result is indeed valid for complex $t$ in the disk $|t-1/2|<1/2$.

The proof is a continuation of the ideas above of Humphries and GH.

We start with summation by parts
$$ \sum_{k=0}^K\mu^k(1-\mu^kt)^n=\frac{(1-\mu^{K+1})(1-\mu^Kt)^n}{1-\mu}-\sum_{k=0}^{K-1}\frac{1-\mu^{k+1}}{1-\mu}[(1-\mu^{k+1}t)^n-(1-\mu^{k}t)^n]$$

$$=\frac{(1-t)^n}{1-\mu}-\frac{\mu^{K+1}(1-\mu^Kt)^n}{1-\mu}+\sum_{k=0}^{K-1}\frac{\mu^{k+1}}{1-\mu}[(1-\mu^{k+1}t)^n-(1-\mu^{k}t)^n]$$

Letting $K\to\infty$ we get

$$f_n(t)=O((1-t)^n)-\frac{\mu}{1-\mu}\sum_{k=0}^\infty\mu^{k}\int_{\mu^{k+1}}^{\mu^k}\frac{d}{dx}(1-xt)^ndx$$

$$=O((1-t)^n)+\frac{n\mu t}{1-\mu}\int_0^1\mu^{-\langle\log_\mu x\rangle}x(1-xt)^{n-1}dx$$

$$=O((1-t)^n)+\frac{n\mu t}{(1-\mu)(n-1)^2}\int_0^{n-1}\mu^{-\langle\log_\mu [s/(n-1)]\rangle}s(1-\frac{ts}{n-1})^{n-1}ds$$

Now, for $t\in(0,1)$, $s\in[0, n-1],$ we obviously have that

$$0\leq 1-\frac{ts}{n-1}\leq e^{-ts/(n-1)}$$ and that there is a constant $M$ independent of $n$ such that $$|e^{-ts/(n-1)}-(1-\frac{ts}{n-1})|\leq \frac{M s^2}{(n-1)^2}.$$

Hence
$$|e^{-ts}-(1-\frac{ts}{n-1})^{n-1}|\leq |e^{-ts/(n-1)}-(1-\frac{ts}{n-1})|(n-1)e^{\frac{-ts(n-2)}{(n-1)}}$$
$$\leq \frac{M s^2e^{-ts/2}}{n-1},\quad n\geq 3,$$

and consequently,

$$\int_0^{n-1}\mu^{-\langle\log_\mu [s/(n-1)]\rangle}s(1-\frac{ts}{n-1})^{n-1}ds=\int_0^{n-1}\mu^{-\langle\log_\mu [s/(n-1)]\rangle}se^{-st}ds+O(1/n)$$

which together with the equality above yields

$$f_n(t)=\frac{n}{(n-1)^2}[\frac{\mu t}{1-\mu}\int_0^{\infty}\mu^{-\langle\log_\mu s-\log_\mu (n-1)\rangle}se^{-st}ds+O(1/n)]$$

and the claim made at the beginning follows.