I'm sorry if the English isn't correct, I'm not native.
Let's say for example these two:
$$\frac{3n}{2n+3}(-1)^{n+1}-\frac{(-1)^n}{2n}$$
$$((-1)^n+1/n)^2$$
I'm sorry if the English isn't correct, I'm not native.
Let's say for example these two:
$$\frac{3n}{2n+3}(-1)^{n+1}-\frac{(-1)^n}{2n}$$
$$((-1)^n+1/n)^2$$
If you just want the bounded property, you can simply find crude upper and lower bounds.
E.g. if $(-1)^n+1/n$ is bounded, then so is $((-1)^n+1/n)^2$ (so we can essentially disregard the power of 2). Then we note that both $(-1)^n$ and $1/n$ are between $-1$ and $1$ for all $n \geq 1$ (I told you it was crude). Hence $-2 \leq (-1)^n+1/n \leq 2$ and $((-1)^n+1/n)^2 \leq 4$.
The first sequence $\frac{3n}{2n+3}(-1)^{n+1}-\frac{(-1)^n}{2n}$ can be resolved by similar techniques.