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This is my third response. I claim that for $0 < t < 1$ we have the uniform bounds $$\liminf_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$ $$\limsup_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$ where $O(1)$ denotes quantities that are absolutely bounded. First of all, $$f_n(t) = \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d[x]$$ $$= \int_{0}^\infty \mu^x(1-\mu^xt)^n\ dx - \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d\langle x\rangle,$$ where $x=[x]+\langle x\rangle$ is the decomposition into integral and fractional parts. Evaluating the first integral on the right, and applying integration parts on the second integral, we obtain $$f_n(t) = \frac{1-(1-t)^{n+1}}{(n+1)(-t\log\mu)}+ (1-t)^n+\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx.$$ Here $$\frac{d}{dx}(\mu^x(1-\mu^xt)^n) = (\log\mu)\mu^x(1-\mu^xt)^{n-1}(1-(n+1)t\mu^x)$$ changes sign only once, namely where $\mu^x=\frac{1}{(n+1)t}$, hence the last integral can be estimated readily by $0\leq \langle x\rangle < 1$ as $$\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx\ll (1-t)^n+\frac{1}{nt}.$$ It follows that $$(n+1)tf_n(t)=\frac{-1}{\log\mu}+O_\mu(n(1-t)^{n})+O(1),$$ which implies the claim above, noting that $\lim_{n\to\infty}n(1-t)^{n}=0$.

Based on the above argument I doubt that $\lim_{n\to\infty}n f_n(t)$ exists. More precisely, I don't think that $$\lim_{n\to\infty} n\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx$$ exists, because the integral is very sensitive on the fractional part of $\log_\mu\frac{1}{(n+1)t}$ where the derivative in the integrand changes its sign.

EDIT: It is easy to verify in an elementary fashion that $\lim_{n\to\infty} nf_n(t)$ does not always exist. Let $t:=\frac{1}{r^2}$ and $\mu:=\frac{1}{r^2}$, where $r>1$ is an integer. Then for any integer $\ell>0$ we have $$n=r^{2\ell} \ \Longrightarrow\ nf_n(t) > n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n=r^2(1-1/n)^n \gg r^2,$$ $$n=r^{2\ell+1}\ \Longrightarrow\ nf_n(t) \ll r^3 2^{-re^{-r} + r \ll r.$$ For the last estimate use that $\mu^x(1-\mu^xt)^n$ increases for $x\leq\ell-\frac{1}{2}$, hence $$n\sum_{k=0}^{\ell-2}\mu^k(1-\mu^kt)^n < n\int_0^{\ell-1}\mu^x(1-\mu^xt)^n\ dx < r^2(1-r^{-2\ell})^{r^{2\ell+1}} < r^2 2^{-r}e^{-r},$$ while clearly $$n\sum_{k=\ell-1}^{\infty}\mu^k(1-\mu^kt)^n < n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n+\frac{n\mu^\ell}{1-\mu} < r^3 2^{-r}+2r.$$ e^{-r}+2r.$$Hence \lim_{n\to\infty} nf_n(t) does not exist when r is sufficiently large. 14 deleted 14 characters in body This is my third response. I claim that for 0 < t < 1 we have the uniform bounds$$ \liminf_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1), \limsup_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$where O(1) denotes quantities that are absolutely bounded. First of all,$$ f_n(t) = \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d[x]  = \int_{0}^\infty \mu^x(1-\mu^xt)^n\ dx - \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d\langle x\rangle,$$where x=[x]+\langle x\rangle is the decomposition into integral and fractional parts. Evaluating the first integral on the right, and applying integration parts on the second integral, we obtain$$ f_n(t) = \frac{1-(1-t)^{n+1}}{(n+1)(-t\log\mu)}+ (1-t)^n+\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx.$$Here$$ \frac{d}{dx}(\mu^x(1-\mu^xt)^n) = (\log\mu)\mu^x(1-\mu^xt)^{n-1}(1-(n+1)t\mu^x) $$changes sign only once, namely where \mu^x=\frac{1}{(n+1)t}, hence the last integral can be estimated readily by 0\leq \langle x\rangle < 1 as$$\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx\ll (1-t)^n+\frac{1}{nt}.$$It follows that$$ (n+1)tf_n(t)=\frac{-1}{\log\mu}+O_\mu(n(1-t)^{n})+O(1), $$which implies the claim above, noting that \lim_{n\to\infty}n(1-t)^{n}=0. Based on the above argument I doubt that \lim_{n\to\infty}n f_n(t) exists. More precisely, I don't think that$$ \lim_{n\to\infty} n\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx $$exists, because the integral is very sensitive on the fractional part of \log_\mu\frac{1}{(n+1)t} where the derivative in the integrand changes its sign. EDIT: It is easy to verify in an elementary fashion that \lim_{n\to\infty} nf_n(t) does not always exist. Let t:=\frac{1}{r^2} and \mu:=\frac{1}{r^2}, where r>1 is an integer. Then for any integer \ell>0 we have$$n=r^{2\ell} \ \Longrightarrow\ nf_n(t) > n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n=r^2(1-1/n)^n \gg r^2,n=r^{2\ell+1}\ \Longrightarrow\ nf_n(t) \ll r^3 2^{-r} + r \ll r.$$For the last estimate use that \mu^x(1-\mu^xt)^n increases for x\leq\ell-1 and decreases for x\geq\ell, x\leq\ell-\frac{1}{2}, hence$$ n\sum_{k=0}^{\ell-2}\mu^k(1-\mu^kt)^n < n\int_0^{\ell-1}\mu^x(1-\mu^xt)^n\ dx < r^2(1-r^{-2\ell})^{r^{2\ell+1}} < r^2 2^{-r} 2^{-r}, $$and also while clearly$$ n\sum_{k=\ell-1}^{\infty}\mu^k(1-\mu^kt)^n < n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n+\frac{n\mu^\ell}{1-\mu} < r^3 2^{-r}+2r.$$Hence \lim_{n\to\infty} nf_n(t) does not exist when r is sufficiently large. 13 added 341 characters in body; added 5 characters in body This is my third response. I claim that for 0 < t < 1 we have the uniform bounds$$ \liminf_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1), \limsup_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$where O(1) denotes quantities that are absolutely bounded. First of all,$$ f_n(t) = \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d[x]  = \int_{0}^\infty \mu^x(1-\mu^xt)^n\ dx - \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d\langle x\rangle,$$where x=[x]+\langle x\rangle is the decomposition into integral and fractional parts. Evaluating the first integral on the right, and applying integration parts on the second integral, we obtain$$ f_n(t) = \frac{1-(1-t)^{n+1}}{(n+1)(-t\log\mu)}+ (1-t)^n+\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx.$$Here$$ \frac{d}{dx}(\mu^x(1-\mu^xt)^n) = (\log\mu)\mu^x(1-\mu^xt)^{n-1}(1-(n+1)t\mu^x) $$changes sign only once, namely where \mu^x=\frac{1}{(n+1)t}, hence the last integral can be estimated readily by 0\leq \langle x\rangle < 1 as$$\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx\ll (1-t)^n+\frac{1}{nt}.$$It follows that$$ (n+1)tf_n(t)=\frac{-1}{\log\mu}+O_\mu(n(1-t)^{n})+O(1), $$which implies the claim above, noting that \lim_{n\to\infty}n(1-t)^{n}=0. Based on the above argument I doubt that \lim_{n\to\infty}n f_n(t) exists. More precisely, I don't think that$$ \lim_{n\to\infty} n\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx $$exists, because the integral is very sensitive on the fractional part of \log_\mu\frac{1}{(n+1)t} where the derivative in the integrand changes its sign. EDIT: It is easy to verify in an elementary fashion that \lim_{n\to\infty} nf_n(t) does not always exist. Let t:=\frac{1}{r^2} and \mu:=\frac{1}{r^2}, where r>0 r>1 is an integer. Then for any integer \ell>0 we have$$n=r^{2\ell} \ \Longrightarrow\ nf_n(t) > n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n=r^2(1-1/n)^n \gg r^2,n=r^{2\ell+1}\ \Longrightarrow\ nf_n(t) \ll \max_{k=\ell,\ell-1}\ n\mu^k(1-\mu^kt)^n\ll r^3 2^{-r} + r \ll r.$$For the last estimate use that \mu^x(1-\mu^xt)^n increases for x\leq\ell-1 and decreases for x\geq\ell, hence$$ n\sum_{k=0}^{\ell-2}\mu^k(1-\mu^kt)^n < n\int_0^{\ell-1}\mu^x(1-\mu^xt)^n\ dx < r^2(1-r^{-2\ell})^{r^{2\ell+1}} < r^2 2^{-r} $$and also$$ n\sum_{k=\ell-1}^{\infty}\mu^k(1-\mu^kt)^n < n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n+\frac{n\mu^\ell}{1-\mu} < r^3 2^{-r}+2r. Hence $\lim_{n\to\infty} nf_n(t)$ does not exist when $r$ is sufficiently large.

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