limit superior of alternating series

Question

I was asked in an exercise to check whether the following series converges or not: $$\sum_{n=1}^{\infty}\left(1+\frac{(-1)^n-3}{n} \right)^{n^2}$$

I used the fact that if a series $(a_n)_n$ converges, then:

$$\limsup_{n \to \infty} \sqrt[n] a_n<1$$

So I calculated $$\limsup_{n \to \infty} \left(1+\frac{(-1)^n-3}{n} \right)^{n} $$

I got $\frac{1}{e^2}$ but the thing is that, I've never calculated a limit superior before, only regular limits. While evaluating the limit I made the assumption that:

$$\limsup_{n \to \infty} \left(1+\frac{(-1)^n-3}{n} \right)^{n} =\lim_{n \to \infty} \left(1+\frac{1-3}{n} \right)^{n} $$

My question is, is this assumption correct?

$$\limsup_{n \to \infty}(-1)^n a_n = \lim_{n \to \infty} a_n?$$

And the same way:

$$\liminf_{n \to \infty}(-1)^n a_n = \lim_{n \to \infty} -a_n?$$

Answer 1

When you want to calculate a limit superior first find an expression in terms of $n$ for $\sup\ \{a_m:m>n\}$, you can call this sequence $s_n:=\sup\ \{a_m:m>n\}$. And then, when you have that expression, just take the limit.

Answer 2

The series converges by the root test, once we note that $0<\left(1+\frac{(-1)^n-3}{n} \right)^{n^2}\le \left(1+\frac{-2}{n} \right)^{n^2}$ and $\left(1+\frac{-2}{n} \right)^{n}\to e^{-2}<1.$