How do you find the limit of $8+(-1)^n 8$? And the intuitive way of thinking why it approaches the limit. Given $\lim_{n\to \infty} 8+(-1)^n8$ does this converge towards some number or does it diverge? What is the intuitive way of thinking about the limit of this function?
 A: Notice that if $n$ is even, then $(-1)^n=1$ (because $n$ is even) so we have
$$
8+(-1)^n8=8+8=16
$$
But if $n$ is odd then $(-1)^n=-1$ because $n$ is odd and then we have
$$
8+(-1)^n8=8-8=0
$$
If the limit is to exist, then if $n$ is 'large', it must approach a single value. But the value 'bounces' back and forth between $0$ and $16$. Therefore, the limit does not converge, i.e. the limit diverges. 
People new to the idea of convergence typically get the idea that 'diverge' means to get very large or very negative. But diverge just means that a sequence does not converge. To converge, you must get very 'close' to a single number. But there are 'plenty' of sequences which do not approach anything in particular. As another example, $\{(-1)^n\}$ simply alternates between $1$ and $-1$. Hence, this sequence also diverges. The sequence $\{(-1)^nn\}$ alternates between large even numbers and very negative odd numbers and hence also diverges (and does not always get 'large' or 'very' negative). You can also come up with sequences that do not exhibit this effect. For instance, $\{\sin n\}$ does not approach any particular value, hence diverges. As does the sequence $\{1,\dfrac{1}{1},2,\dfrac{1}{2},3,\dfrac{1}{3},4,\dfrac{1}{4},\ldots\}$. Come up with lots more on your own!
