Classify whether the limit converges or diverges and if its bounded or unbounded $$\left\{\frac{(-1)^n100^n}{n!}\right\}$$
So, I know $\left\{\frac{100^n}{n!}\right\}$ is a basic null sequence, although here itll oscillate between negative and positive values dependent on the odd/even nature of $n$. However, I think it'll still converge to $0$ and be a null sequence eventually. and its also bounded. Am I wrong?
 A: Squeeze theorem will do
$$\frac{-100^n}{n!}\leq\frac{(-1)^n100^n}{n!}\leq \frac{100^n}{n!}$$
Both of which converge to zero.
A: By the formal definition of limit, we say that $a_n \to 0$ if given any $\epsilon >0$ there exist $n_0 \in \Bbb{N} $ such that $|a_n|<\epsilon$ for all $n\geq n_0$. Since we are only concerned about the absolut value, the $(-1)^n $ doesn't really matter in this case. Therefore since $\frac {100^n}{n!} \to 0$ also does $\frac {(-1)^n 100^n}{n!}$
A: Notice that if $|a_n| \to 0$, then $a_n \to 0$.
Suppose that $a_n \to 0$. Then there exists $N \in \mathbb N$ so that $|a_N|<\epsilon$ for any $\epsilon$.  Then the result is clear. Notice that you need the limit to be zero. For example, $\{(-1)^n\}_{n=1}^{\infty}$ does not converge, while its absolute value clearly does. 
Anyway, apply the first fact to solve your problem.
edit every convergent sequence is bounded. Let $\epsilon>0$, and $a_n$ be a sequence that converges to $L$. There are two options:
There exists some $N$ so that $n \geq N$ implies that $|a_n-L|<\epsilon$, and so $a_n<L+\epsilon$.
Otherwise, take $\max\{a_1,\ldots,a_{N-1}\}=M_1$.
Now take $\max\{L+\epsilon, M_1\}$. Clearly this bounds the sequence$(a_n)$
