Pointwise convergence of $f_n(x) = \left(\frac{3^{x/n} + e^{x/n}}{2} \right)^{-n}$

Question

I am given the sequence $f_n : [0,\infty) \to \mathbb R$ such that $$ f_n(x) = \left(\frac{3^{x/n} + e^{x/n}}{2} \right)^{-n} $$ and I am asked to discuss its convergence almost everywhere w.r.t. the standard Lebesgue measure on $[0,\infty)$ and find its limiting function. I have found two ways to rewrite $f_n(x)$: one using the binomial theorem, $$ f_n(x) = \left[ \frac{e^x}{2^n} \sum_{k=0}^n \binom n k e^{\frac k n (\ln 3 - 1) x} \right]^{-1}; $$ the other by expanding the exponentials into their respective Maclaurin series, $$ f_n(x) = \left[ \sum_{k=0}^\infty \frac{1-\ln^k3}{2 n^k k!} x^k \right]^{-n}; $$ however, neither of these rewritings have helped me come closer to finding $f = \lim_{n\to\infty} f_n$. Any hints?

Answer 1

Consider the quantity $g_n=-\log f_n$ given by $$ g_n=n\log\frac{3^{x/n}+e^{x/n}}{2}. $$ Let $\epsilon=1/n$. By L'Hospital's rule, $$ \lim_{n\to\infty}g_n=\lim_{\epsilon\to 0}\frac{\log(3^{\epsilon x}+e^{\epsilon x})-\log 2}{\epsilon}=\lim_{\epsilon\to 0}\frac{x\log 3\cdot 3^{\epsilon x}+x e^{\epsilon x}}{3^{\epsilon x}+e^{\epsilon x}}=x\frac{\log 3 + 1}{2}. $$ Thus, the limit $f$ is given by $$ f=e^{-x\alpha},\qquad \alpha=\frac{\log 3 + 1}{2}. $$ As a side note, observe that $f$ is obtained by taking the geometric mean of the two terms in $f_n$ instead of the arithmetic mean.

Answer 2

Hint: For $n\gg x$, we have $$ \frac{3^{x/n}+e^{x/n}}{2}=1+\frac{x(1+\log3)}{2n}+o(n^{-1}). $$

Answer 3

Apply L'Hopital's Rule to show that $\frac 1 y \log(\frac {e^{xy}+e^{xy}} 2)$ tends to $\frac {x\log \, 3+x} 2$ as $ y \to 0$. This gives the required limit as $e ^{-\frac {x\log \, 3+x} 2}$.