Límite del valor mínimo de una integral

Question

Que $$f(a)=\frac{1}{2}\int_{0}^{1}|ax^n-1|dx+\frac{1}{2}$$ Here $n $ is a natural number. Let $b_n $ be the minimum value of $f (a) $ for $a > 1$. Evaluate $% $ $\lim_{m \to \infty}b_mb_{m+1}\ldots b_{2m}$

Algunos entrantes por favor. Gracias.

Answer 1

$$\begin{eqnarray*}f(a) = \frac{a}{2}\int_{0}^{1}\left| x^n-\frac{1}{a}\right|\,dx+\frac{1}{2}&=&\frac{1}{2}+\frac{a}{2}\int_{0}^{1}(x^n-1/a)\,dx+a\int_{0}^{\frac{1}{\sqrt[n]{a}}}\left(\frac{1}{a}-x^n\right)\,dx\\&=&\frac{1}{2}+\frac{a}{2n+2}-\frac{1}{2}+\frac{1}{\sqrt[n]{a}}-\frac{1}{(n+1)\sqrt[n]{a}}\\&=&\frac{a}{2n+2}+\frac{n}{(n+1)\sqrt[n]{a}}\end{eqnarray*}$ $ alcanza su mínimo en $a=2^{\frac{n}{n+1}}$: $$ b_n = 2^{-\frac{1}{n+1}}.$ $ entonces considerar que: $$ \sum_{k=m}^{2m}\frac{1}{k+1}\stackrel{m\to +\infty}{\longrightarrow}\log 2.$ $