$\left\{x_n\right\}$ is a sequence where $x_n=\frac{(-1)^n}{n}+\sin\frac{n\pi}{2}$. Find $\sup\left\{x_n\right\}$ and $\inf\left\{x_n\right\}$.
It is a question in my textbook. The answer is: $\sup\left\{x_n\right\}=1$ and $\inf\left\{x_n\right\}=-1-\frac{1}{3}$. I cannot get to this answer. I proceed like this:
$\frac{(-1)^n}{n}=-1, -1/3, -1/5,...$(when $n$ is odd), $1/2, 1/4, 1/6,...$(when $n$ is even).
$\sin\frac{n\pi}{2}=1, -1$(when $n$ is odd), $0$(when $n$ is even).
So, $\sup\left\{x_n\right\}=\frac{1}{2}$ and $\inf\left\{x_n\right\}=-2$.
Where is the mistake? How can I find the correct answer? Does $\left\{x_n\right\}$ converge? Please anyone help me solving this. Thanks in advance.
Your conclusion about $\sin(n\pi/2)$ is not quite correct. Maybe just write a little more detail and see what happens: $$x_{4n\pm1} = \dfrac{-1}{4n\pm 1}\pm 1,\, x_{4n} = \dfrac{1}{4n}+1$$ and $$x_{4n+2} = \dfrac{1}{4n+2} - 1.$$
Check that above holds and consider them separately.