Solved but verifying my answer on divergent sequenues.

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 :)

Re: Solved but verifying my answer on divergent sequenues.

Hey rahul.

A simple test would be to use the ratio test and show that the ratio is greater than 1 in absolute value. Are you familiar with this test?

Re: Solved but verifying my answer on divergent sequenues.

yeah i am familiar with the test, one has to take the ratio of n+1 term to the nth term in absolute value and if it is greater than 1 it is divergent, less than 1 then convergent that kind of stuff right? i haven't done that yet but i know. Anyways i jus wanted to check that what i did was right or wrong? Also can u tell me how do i prove (-1)^n times n squared is not convergent. just give me some hints, i'll do the rest.

Thank you.