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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 $$n=r^{2\ell} \ \Longrightarrow\ nf_n(t) > n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n=r^2(1-1/n)^n \gg r^2,$$ $$n=r^{2\ell+1}\ \Longrightarrow\ nf_n(t) \ll r^3 2^{-r} + r \ll r.$$ For the last estimate use that $\mu^x(1-\mu^xt)^n$ increases for $x\leq\ell-\frac{1}{2}$, hence $$ n\sum_{k=0}^{\ell-2}\mu^k(1-\mu^kt)^n < n\int_0^{\ell-1}\mu^x(1-\mu^xt)^n\ dx < r^2(1-r^{-2\ell})^{r^{2\ell+1}} < r^2 2^{-r}, $$ 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.

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 $$n=r^{2\ell} \ \Longrightarrow\ nf_n(t) > n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n=r^2(1-1/n)^n \gg r^2,$$ $$n=r^{2\ell+1}\ \Longrightarrow\ nf_n(t) \ll r^3 e^{-r} + r \ll r.$$ For the last estimate use that $\mu^x(1-\mu^xt)^n$ increases for $x\leq\ell-\frac{1}{2}$, hence $$ n\sum_{k=0}^{\ell-2}\mu^k(1-\mu^kt)^n < n\int_0^{\ell-1}\mu^x(1-\mu^xt)^n\ dx < r^2(1-r^{-2\ell})^{r^{2\ell+1}} < r^2 e^{-r}, $$ while clearly $$ n\sum_{k=\ell-1}^{\infty}\mu^k(1-\mu^kt)^n < n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n+\frac{n\mu^\ell}{1-\mu} < r^3 e^{-r}+2r.$$ Hence $\lim_{n\to\infty} nf_n(t)$ does not exist when $r$ is sufficiently large.

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 $x\geq\ell$, 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.

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 $x\leq\ell-\frac{1}{2}$, hence $$ n\sum_{k=0}^{\ell-2}\mu^k(1-\mu^kt)^n < n\int_0^{\ell-1}\mu^x(1-\mu^xt)^n\ dx < r^2(1-r^{-2\ell})^{r^{2\ell+1}} < r^2 2^{-r}, $$ 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.

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 $r>0$ is an integer. Then for any integer $\ell>0$ we have $$n=r^{2\ell} \ \Longrightarrow\ nf_n(t) > n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n=r^2(1-1/n)^n \gg r^2,$$ $$n=r^{2\ell+1}\ \Longrightarrow\ nf_n(t) \ll \max_{k=\ell,\ell-1}\ n\mu^k(1-\mu^kt)^n\ll r^3 2^{-r} + r \ll r.$$ Hence $\lim_{n\to\infty} nf_n(t)$ does not exist when $r$ is sufficiently large.

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 $x\geq\ell$, 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.