0
$\begingroup$

I need Help evaluating the limit of $$\lim_{n \to \infty} \left| \cos \left( \frac{\pi}{4(n-1)} \right) \right|^{2n-1} = L$$

I already know that $L = 1$, but I need help showing it.

The best idea I could come up with was to take the series representation of cosine. $$\lim_{j,n \to \infty} \left| 1 - \left( \frac{\pi}{4(n-1)} \right)^2 \frac{1}{2!} + ....+ \frac{(-1)^j}{(2j)!}\left( \frac{\pi}{4(n-1)} \right)^{2j} \right|^{2n-1} = L$$

All lower order terms go to zero leaving:

$$\lim_{j,n \to \infty} \left|\frac{1}{(2j)!}\left( \frac{\pi}{4(n-1)} \right)^{2j} \right|^{2n-1} = L$$

But this doesn't really seem like I am any closer. How do I proceed? Obviously L'Hospitals rule will occur eventually. Hints?

$\endgroup$
2
$\begingroup$

Put $$L = \lim_{n \to \infty} \left| \cos \left( \frac{\pi}{4(n-1)} \right) \right|^{2n-1}$$ Then $$\log(L) = \lim_{n \to \infty} (2n-1)\log\left( \cos \left( \frac{\pi}{4(n-1)} \right) \right).$$ This is an $0\cdot \infty$ indeterminate form. Put the $2n -1$ in the denominator and invoke L'hospital.

$\endgroup$
0
$\begingroup$

Your idea of using, for a sufficiently large value of $n$, different Taylor expansions looks good to me. Built around $x=0$, we have $\cos(x)=1-\frac{x^2}{2}+O\left(x^3\right)$. On the other hand, $(1-y)^k$ is almost $(1-k y)$. So, replace $y$ by $x^2$, $k$ by $(2n-1)$,$x$ by $\frac{\pi }{4 (n-1)}$ and you end with $$1-\frac{\pi ^2 (2 n-1)}{32 (n-1)^2}$$ which, for a sufficiently large value of $n$ is almost $$1-\frac{\pi ^2}{16 n}$$Now, push $n$ to infinity.

$\endgroup$

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.