I was asked to show that 2^{n}is divergent. Here is what i did, i assumed that it was convergent and used the fact that a convergent sequence needs to be bounded. let M be the upper bound ,then 2^{n}< M for all natural numbers, then this implies that n < logM/log2 for all natural numbers, which contradicts the Archimedian property. So is this right?

Also i am asked to give two divergent sequences whose product is convergent, i saw an example in this very own site (-1)^{n}and (-1)^{n+1}are two divergent sequences whose product is convergent. Can u give me some more examples?

Thank you