1
$\begingroup$

I have a sequence:

$$ X_n\:=\:\cos\left(\left(\frac{3^n+\pi^n}{3^n+\left(\pi-\frac{1}{4} \right)^n} \right)^{1/n}\right) $$

I have to find the limit when $n \to \infty $ where $n \in \mathbb{N}$.

Which is the best way to find the answer ? Can I reduce or use the Squeeze theorem in that case ?

$\endgroup$
  • 1
    $\begingroup$ Almost as a reflex, I'd go for Taylor expansions. $\endgroup$ – Clement C. Aug 28 '17 at 18:07
  • $\begingroup$ So far no one has up-voted the question among the five people who have who have answered it. $\endgroup$ – Michael Hardy Aug 28 '17 at 18:53
  • $\begingroup$ It seems as if the cosine is tacked on to the question as an afterthought. If one finds the limit of the expression inside the cosine function, then evaluating the cosine function at that point is a separate issue. (Although a priori it's not inconceivable that the sequence inside the cosine function has a bunch of subsequential limits with in the set $\{a + 2\pi k: k\in\mathbb Z\},$ so the limit of that inside sequence would not exist but the limit of its cosine would. $\endgroup$ – Michael Hardy Aug 28 '17 at 19:06
4
$\begingroup$

$${3^n+\pi^n\over3^n+(\pi-{1\over4})^n}=\left(\pi\over3\right)^n{1+\left(3\over\pi\right)^n\over1 +\left(\pi-{1\over4}\over3\right)^n}$$

Now $\pi-{1\over4}\lt3\lt\pi$, so $1+(3/\pi)^n\to1$ and $1+((\pi-{1\over4})/3)^n\to1$. Thus

$$\left(3^n+\pi^n\over3^n+(\pi-{1\over4})^n\right)^{1/n}\to{\pi\over3}$$

Taking the cosine gives the limit $\cos(\pi/3)=1/2$.

$\endgroup$
  • $\begingroup$ My first step was going to be $$ \left( \frac{3^n+\pi^n}{3^n+\left(\pi-\frac{1}{4} \right)^n} \right)^{1/n} = \frac{ \left( 1 + \left( \frac \pi 3 \right)^n \right)^{1/n}}{\left( 1 + \left( \frac{4\pi-1}{12} \right)^n \right)^{1/n}}. $$ Maybe you've already done what I was going to do. $\endgroup$ – Michael Hardy Aug 28 '17 at 18:32
  • 1
    $\begingroup$ @MichaelHardy, I thought it was a little more clear to pull the dominant terms out of the numerator and denominator, leaving stuff that tended to obvious limits. $\endgroup$ – Barry Cipra Aug 28 '17 at 20:09
0
$\begingroup$

The answer is $\frac12$, because$$\sqrt[n]{\frac{3^n+\pi^n}{3^n+\left(\pi-\frac14\right)^n}}$$behaves as$$\sqrt[n]{\frac{\pi^n}{3^n}},$$that is, as $\frac\pi3$, when $n$ is very large. And $\cos\left(\frac\pi3\right)=\frac12$.

$\endgroup$
0
$\begingroup$

Almost as a reflex, I'd go for Taylor expansions. (Because it works. Not always the most elegant, bu at least it gets us somewhere.) When $n\to\infty$, letting $x\stackrel{\rm def}{=} \frac{\pi -\frac{1}{4}}{3}$ and $y\stackrel{\rm def}{=} \frac{3}{\pi}$, with $x>y$ (so that $y^n=o(x^n)$)

\begin{align} \cos\left(\left(\frac{3^n+\pi ^n}{3^n+\left(\pi -\frac{1}{4}\right)^n}\right)^{\frac{1}{n}}\right) &= \cos\left(\frac{\pi}{3}\left(\frac{1+\left(\frac{3}{\pi}\right)^n}{1+\left(\frac{\pi -\frac{1}{4}}{3}\right)^n}\right)^{\frac{1}{n}}\right)\\ &= \cos\left(\frac{\pi}{3}\left(\left(1+y^n\right)\left(1+x^n + o(x^{n})\right)\right)^{\frac{1}{n}}\right)\\ &= \cos\left(\frac{\pi}{3}\left(1+y^n + x^n +o(x^n)\right)^{\frac{1}{n}}\right) \\ &= \cos\left(\frac{\pi}{3}\left(1+x^n +o(x^n)\right)^{\frac{1}{n}}\right) \tag{as $x>y$}\\ &= \cos\left(\frac{\pi}{3}\exp\left(\frac{1}{n}\ln\left(1+x^n +o(x^n)\right)\right)\right) \\ &= \cos\left(\frac{\pi}{3}\exp\left(\frac{x^n}{n}+o\left(\frac{x^n}{n}\right)\right)\right) \\ &= \cos\left(\frac{\pi}{3}\left(1+\frac{x^n}{n}+o\left(\frac{x^n}{n}\right)\right)\right) \\ &\xrightarrow[n\to\infty]{} \cos\frac{\pi}{3} = \frac{1}{2} \end{align}

$\endgroup$
0
$\begingroup$

HINT: write your term as $$\cos\left(\frac{1+\left(\frac{\pi}{3}\right)^n}{1+\left(\frac{\pi-\frac{1}{4}}{3}\right)^n}\right)^{1/n}$$

$\endgroup$
0
$\begingroup$

Denote the argument of the cosine as $a_n$. Estimate: $$\left( \frac{\pi ^n \left(\left(\frac{3}{\pi}\right)^n+1\right)}{2\cdot 3^n}\right)^{1/n}<a_n<\left(\frac{2\cdot \pi ^n}{3^n \left(1+\left(\frac{\pi -\frac14}{3}\right)^n\right)}\right)^{1/n}.$$ Taking limit: $$\frac{\pi }{3}\le \lim_{n\to\infty} a_n \le \frac{\pi}{3}.$$ Hence: $$\lim_{n\to\infty} \cos{a_n} =\cos{\frac{\pi}{3}}=\frac12.$$

$\endgroup$
0
$\begingroup$

$$a_n=\sqrt[n]{\frac{3^n+ \pi^n}{3^n+ (\pi -1/4)^n}}\leq \sqrt[n]{\frac{2 \pi^n}{3^n}}=\sqrt[n]{2} \frac{\pi}{3} \to \frac{\pi}{3}$$

Also $$a_n=\sqrt[n]{\frac{3^n+ \pi^n}{3^n+ (\pi -1/4)^n}} \geq \sqrt[n]{ \frac{\pi^n}{3^n+3^n}}=\frac{1}{\sqrt[n]{2}} \frac{\pi}{3} \to \frac{\pi}{3}$$

So from squeeze theorem $a_n \to \frac{\pi}{3}$.

Now using the continuity of $\cos{x}$ we have that $$\cos{a_n} \to \cos{\frac{\pi}{3}}=\frac{1}{2}$$

$\endgroup$

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.