5
$\begingroup$

The problem I am stuck on asks the reader to find the following limit: $$\lim_{n \rightarrow \infty} \int^{n}_{0} \left(1+\frac{x}{n}\right)^{-n} \log\left(2+ \cos\left(\frac{x}{n}\right)\right)\ \mathrm dx.$$ The section I am working on contains all your basic limit theorem in measure theory (Monotone Covergence Theorem, Fatou's Lemma, Dominated Convergence Theorem). I know I am probably overseeing an application of one of them. Help would be greatly appreciated.

$\endgroup$
  • 1
    $\begingroup$ Domainated convergence theorem is your tool. $\endgroup$ – Mhenni Benghorbal Jul 18 '14 at 22:43
3
$\begingroup$

For $0 \leq y \leq 1$ we have $\log(1+y) \geq y - y^2/2$, so for $0 < x \leq n$ we have

$$ \begin{align} \left(1+\frac{x}{n}\right)^{-n} &= \exp\left[ -n \log\left(1+\frac{x}{n}\right)\right] \\ &\leq \exp\left[-n \left(\frac{x}{n} - \frac{x^2}{2n^2}\right)\right] \\ &= \exp\left[-x + \frac{x^2}{2n}\right] \\ &\leq \exp\left[-x + \frac{x^2}{2x}\right] \\ &= e^{-x/2}. \end{align} $$

Now apply dominated convergence.

$\endgroup$
2
$\begingroup$

hints:

$$\ln(2+\cos(x/n)) \leq \ln(3).$$

$\endgroup$
  • $\begingroup$ For the second term $\log (2+...) $ you may use the inequality $$\frac {x}{1+x} < \log (1+x)<x $$ valid for all, $ x>-1$ with $ x\ne 0$. $\endgroup$ – mwomath Jul 19 '14 at 3:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.