Sup and inf of a sequence

Question

I have to find $\inf$ and $\sup$ of the given sequence: $$ a_n := \left(1+ \frac{\cos(n\pi)}{n}\right)^n + \sin\left(\frac{n\pi}{2}\right)$$ First thought that come to my mind was to consider 4 cases dependinig on trigonometric functions. For $\sup$, 2 cases, when $\sin = 1$ or $\cos = 1$, and the same for $\inf$.
Is that correct or I'm completely wrong?

Answer 1

We assume $n$ must be a positive integer.

We will split into cases depending on the remainder of $n$ when divided by $4$. We will use the fact that $f(x) = \big(1 + \frac{m}{x}\big)^{x}$ is increasing on $x\in (1,\infty)$ for $m \geq -1$, which can be proven using basic calculus. We will also use $\displaystyle\lim_{x\to\infty}f(x) = e^{m}$.

If $n = 4k$ for some integer $k$:

$$a_{4k} = \bigg(1 + \frac{\cos(4k\pi)}{4k}\bigg)^{4k} + \sin(2\pi k)=\bigg(1 + \frac{1}{n}\bigg)^{n}$$

This sequence is increasing, so its infimum is $a_{4} =\frac{625}{256}$ and its supremum is $\displaystyle\lim_{k\to\infty}a_{4k} = e$.

If $n = 4k + 1$:

$$a_{4k + 1} = \bigg(1 + \frac{\cos(4\pi k + \pi)}{4k + 1}\bigg)^{4k + 1} + \sin\bigg(2\pi k + \frac{\pi}{2}\bigg) = \bigg(1-\frac{1}{n}\bigg)^{n} + 1$$

This sequence is increasing, so its infimum is $a_{1} = 1$ and its supremum is $\displaystyle\lim_{k\to\infty}a_{4k + 1} = \frac{1}{e} + 1$.

If $n = 4k + 2$:

$$a_{4k + 2} = \bigg(1 + \frac{\cos(4k\pi + 2\pi)}{4k + 2}\bigg)^{4k + 2} + \sin(2\pi k + \pi)=\bigg(1 + \frac{1}{n}\bigg)^{n}$$

This sequence is increasing, so its infimum is $a_{2} =\frac{9}{4}$ and its supremum is $\displaystyle\lim_{k\to\infty}a_{4k + 2} = e$.

If $n = 4k + 3$:

$$a_{4k + 3} = \bigg(1 + \frac{\cos(4\pi k + 3\pi)}{4k + 3}\bigg)^{4k + 3} + \sin\bigg(2\pi k + \frac{3\pi}{2}\bigg) = \bigg(1-\frac{1}{n}\bigg)^{n} - 1$$

This sequence is increasing, so its infimum is $a_{3} = -\frac{19}{27}$ and its supremum is $\displaystyle\lim_{k\to\infty}a_{4k + 3} = \frac{1}{e} - 1$.

Then, $\inf a_{n}$ is the minimum of the infima of the subsequences making up $a_{n}$, Likewise, $\sup a_{n}$ is the maximum of the suprema of these subsequences. Thus, $\boxed{\inf a_{n} = -\frac{19}{27}\text{ and }\sup a_{n} = e.}$

Answer 2

Yes, that seems a good idea. Notice that $\cos(n\pi)$ takes values $\pm 1$, while $\sin(\frac{n\pi}{2})$ can also be $0$ (for $n$ even).

You can study the behaviour of the sequence splitting it in the four cases corresponding to $n\ (\operatorname{mod} 4)$.