-1

having problems with this question. not sure where to start. Dont understand how is the monotone convergence theorem s helpful in here

ann
  • 15

1 Answers1

0

We want to show that the sequence $(f_n(x))_{n\in\mathbb{N}}$ converges for all $x\in[0,1]$.

To do this we first of all note that all the terms $2^{-i}(1+\sin(2^i x))$ in the series are nonnegative (since $-1\leq\sin(x)\leq 1$). So the sequence of $f_n(x)$ is monotonically increasing (for all $x$).

Now since $|1+\sin(2^i x)|\leq 2$ for all $i$ and $x$, we see that: \begin{equation} \sum_{i=0}^n 2^{-i}(1+\sin(2^i x))\leq 2\sum_{i=0}^n 2^{-i} \end{equation} Now the last series $\sum_{i=0}^n 2^{-i}$ converges to $2$ as $n\to\infty$, hence the sequence $(f_n(x))_{n\in\mathbb{N}}$ is bounded above by 4. By the monotone convergence theorem, $(f_n(x))_{n\in\mathbb{N}}$ converges (with limit $\leq 4$).

Uncountable
  • 3,520
  • a) ii) all the time we had to prove that function is Lipschitz for very simple functions, but now I've got a series. That just confused me. – ann Feb 16 '15 at 23:07