I was asked in an exercise to check whether a series converges or not and while doing so I got the following limit to solve:
$$\limsup_{n \to \infty} \left(1+\frac{(-1)^n-3}{n} \right)^{n}$$
I did the following in order so solve it:
$$\limsup_{n \to \infty} \left(1+\frac{(-1)^n-3}{n} \right)^{n} =\lim_{n \to \infty} \left(1+\frac{1-3}{n} \right)^{n} =\frac{1}{e^2}$$
The series that I was trying to check if it was convergent or not turned out to be convergent and that implies that:
$$\limsup_{n \to \infty} \left(1+\frac{(-1)^n-3}{n} \right)^{n} < 1$$ Which is what I got. But the thing is that this is the first time that I tried to calculate a Limit superior. So is the following step correct or is it wrong and I just got lucky and got the same result:
$$\limsup_{n \to \infty} \left(1+\frac{(-1)^n-3}{n} \right)^{n} =\lim_{n \to \infty} \left(1+\frac{1-3}{n} \right)^{n}$$
$$\lim_{n \to \infty} \left(1+\frac{(-1)^{2n}-3}{2n} \right)^{2n} = \lim_{n \to \infty} \left(1-\frac{1}{n} \right)^{2n} = \frac{1}{e^2}$$ And $$\lim_{n \to \infty} \left(1+\frac{(-1)^{2n+1}-3}{2n+1} \right)^{2n+1} =\lim_{n \to \infty} \left(1-\frac{4}{2n+1} \right)^{2n+1} = \frac{1}{e^4}$$
Since any subsequence will have infinitely many even or infinitely odd terms, those are the only two possible limit points, so the limsup is the first one.