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Using partial summation, we have that $$\sum_{k = 0}^{K}{\mu^k(1 - \mu^k t)^n} = \frac{(1 - \mu^{K + 1})(1 - \mu^K t)^n}{1 - \mu} - \frac{nt \log \mu^{-1}}{1 - \mu} \int_{0}^{K}{\mu^x (1 - \mu^{\lfloor x \rfloor + 1}) (1 - \mu^x t)^{n-1} \: dx},$$ where $\lfloor x \rfloor$ is the integer part of $x$. By taking the limit as $K$ tends to infinity, $$\sum_{k = 0}^{\infty}{\mu^k(1 - \mu^k t)^n} = \frac{1}{1 - \mu} - \frac{nt \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^x (1 - \mu^{\lfloor x \rfloor + 1}) (1 - \mu^x t)^{n-1} \: dx}.$$ Now a simple calculation shows that $$\frac{1}{1 - \mu} - \frac{nt \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^x (1 - \mu^x t)^{n-1} \: dx} = \frac{(1 - t)^n}{1 - \mu}$$ by making the substitution $u = 1 - \mu^x t$. So the tricky part is the other bit of the integral, which is $$E = \frac{nt \mu \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^{x + \lfloor x \rfloor} (1 - \mu^x t)^{n-1} \: dx}.$$ Note that as $x - 1 < \lfloor x \rfloor \leq x$, we have the bounds $$A \leq E \leq \frac{1}{\mu} A$$ with $$A = \frac{nt \mu \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^{2x} (1 - \mu^x t)^{n-1} \: dx}.$$ Once again, this isn't tricky to calculate: making the same substitution as earlier, we find that $$A = \frac{\mu}{t (1 - \mu)} \frac{1 - (nt + 1) (1 - t)^n}{n + 1}.$$

So piecing everything together, we obtain $$\sum_{k = 0}^{\infty}{\mu^k(1 - \mu^k t)^n} = \frac{(1 - t)^n}{1 - \mu} + E$$ with $$E \asymp_{\mu} \frac{1 - (nt + 1) (1 - t)^n}{t(n + 1)}.$$ This doesn't yield a closed form for $\lim_{n \to \infty} n f_n(t)$, unfortunately, but it does show that $$\frac{\mu}{t(1 - \mu)} \leq \lim_{n liminf_{n \to \infty} n f_n(t) \leq \limsup_{n \to \infty} n f_n(t) \leq \frac{1}{t (1 - \mu)}.$$

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Using partial summation, we have that $$\sum_{k = 0}^{K}{\mu^k(1 - \mu^k t)^n} = \frac{(1 - \mu^{K + 1})(1 - \mu^K t)^n}{1 - \mu} - \frac{nt \log \mu^{-1}}{1 - \mu} \int_{0}^{K}{\mu^x (1 - \mu^{\lfloor x \rfloor + 1}) (1 - \mu^x t)^{n-1} \: dx},$$ where $\lfloor x \rfloor$ is the integer part of $x$. By taking the limit as $K$ tends to infinity, $$\sum_{k = 0}^{\infty}{\mu^k(1 - \mu^k t)^n} = \frac{1}{1 - \mu} - \frac{nt \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^x (1 - \mu^{\lfloor x \rfloor + 1}) (1 - \mu^x t)^{n-1} \: dx}.$$ Now a simple calculation shows that $$\frac{1}{1 - \mu} - \frac{nt \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^x (1 - \mu^x t)^{n-1} \: dx} = \frac{(1 - t)^n}{1 - \mu}$$ by making the substitution $u = 1 - \mu^x t$. So the tricky part is the other bit of the integral, which is $$E = \frac{nt \mu \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^{x + \lfloor x \rfloor} (1 - \mu^x t)^{n-1} \: dx}.$$ Note that as $x - 1 < \lfloor x \rfloor \leq x$, we have the bounds $$A \leq E \leq \frac{1}{\mu} A$$ with $$A = \frac{nt \mu \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^{2x} (1 - \mu^x t)^{n-1} \: dx}.$$ Once again, this isn't tricky to calculate: making the same substitution as earlier, we find that $$A = \frac{(n + t) frac{\mu}{t (1 - t)^n \mu)} \frac{1 - 1}{(n (nt + 1) t (1 - \mu)}.$$t)^n}{n + 1}.$$So piecing everything together, we obtain$$\sum_{k = 0}^{\infty}{\mu^k(1 - \mu^k t)^n} = \frac{(1 - t)^n}{1 - \mu} + E$$with$$E \asymp_{\mu} \frac{(n frac{1 - (nt + t1) (1 - t)^n - 1}{(n t)^n}{t(n + 1) t}.$$1)}.$$ This doesn't yield a closed form for $\lim_{n \to \infty} n f_n(t)$, unfortunately, but it does show that $$- \frac{1}{t \mu (1 \frac{\mu}{t(1 - \mu)} \leq \lim_{n \to \infty} n f_n(t) \leq - \frac{1}{t (1 - \mu)}.$$

Using partial summation, we have that $$\sum_{k = 0}^{K}{\mu^k(1 - \mu^k t)^n} = \frac{(1 - \mu^{K + 1})(1 - \mu^K t)^n}{1 - \mu} - \frac{nt \log \mu^{-1}}{1 - \mu} \int_{0}^{K}{\mu^x (1 - \mu^{\lfloor x \rfloor + 1}) (1 - \mu^x t)^{n-1} \: dx},$$ where $\lfloor x \rfloor$ is the integer part of $x$. By taking the limit as $K$ tends to infinity, $$\sum_{k = 0}^{\infty}{\mu^k(1 - \mu^k t)^n} = \frac{1}{1 - \mu} - \frac{nt \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^x (1 - \mu^{\lfloor x \rfloor + 1}) (1 - \mu^x t)^{n-1} \: dx}.$$ Now a simple calculation shows that $$\frac{1}{1 - \mu} - \frac{nt \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^x (1 - \mu^x t)^{n-1} \: dx} = \frac{(1 - t)^n}{1 - \mu}$$ by making the substitution $u = 1 - \mu^x t$. So the tricky part is the other bit of the integral, which is $$E = \frac{nt \mu \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^{x + \lfloor x \rfloor} (1 - \mu^x t)^{n-1} \: dx}.$$ Note that as $x - 1 < \lfloor x \rfloor \leq x$, we have the bounds $$A \leq E \leq \frac{1}{\mu} A$$ with $$A = \frac{nt \mu \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^{2x} (1 - \mu^x t)^{n-1} \: dx}.$$ Once again, this isn't tricky to calculate: making the same substitution as earlier, we find that $$A = \frac{(n + t) (1 - t)^n - 1}{(n + 1) t (1 - \mu)}.$$
So piecing everything together, we obtain $$\sum_{k = 0}^{\infty}{\mu^k(1 - \mu^k t)^n} = \frac{(1 - t)^n}{1 - \mu} + E$$ with $$E \asymp_{\mu} \frac{(n + t) (1 - t)^n - 1}{(n + 1) t}.$$ This doesn't yield a closed form for $\lim_{n \to \infty} n f_n(t)$, unfortunately, but it does show that $$- \frac{1}{t \mu (1 - \mu)} \leq \lim_{n \to \infty} n f_n(t) \leq - \frac{1}{t (1 - \mu)}.$$