Let $f_n(x) = (1 + \cos x)^n$ ,$f_n:[0,1]\to\mathbb{R}$ and let $a_n =\displaystyle\int_0^{\frac{1}{2^n}} f_n(x)\,dx$. Calculate the sequence limit or prove that the limit does not exists.
I tried to compute the integral using riemann sum but I got stocked choosing random ci. Can some one help with this question please ?
This should work in an easier way. First substitute $y := 2^nx$. Then $$ a_n = \int^1_0 \left(\frac{1 + \cos(2^{-n}y)}{2} \right)^n~\mathrm{d}y. $$ Now consider $$ b_n(y) := \left(\frac{1 + \cos(2^{-n}y)}{2} \right)^n $$ and note $$ \sup_{y \in [0, 1]} \lvert b_n(y) - 1 \rvert = \sup_{y \in [0, 1]} 1 - b_n(y) = 1 - b_n(1) $$ because $y \mapsto \cos(2^ny)$ is decreasing and $t \mapsto t^n$ is increasing on $[0, 1]$. Then convince yourself that $\displaystyle \lim_{n \rightarrow \infty} b_n(1) = 1$ by writing $$ \lim_{n \rightarrow \infty} b_n(1) = \exp\left(\lim_{n \rightarrow \infty} n\log\left( \frac{1 + \cos(2^{-n}y)}{2}\right)\right) $$ and using l'Hopitals rule.
This proves that $b_n$ converges to $1$ uniformly and therefore limits and integrals can be interchanged. $$ \lim_{n \rightarrow \infty} a_n =\lim_{n \rightarrow \infty} \int^1_0 b_n(y) ~\mathrm{d}y = \int^1_0 \lim_{n \rightarrow \infty} b_n(y) ~\mathrm{d}y = \int^1_0 1~\mathrm{d}y = 1 $$