Sequences

I can tell you now the first one converges by p-series. (p = 2 > 1). Sec.

It's not a true p-series. The numerator is increasing at a constant rate, so it's almost like:

an = n/n^2 = 1/n

which fails the p-series.

3. an = ln (n+1) - ln n
First, I set f(x) = y = ln ((x+1)/x). Then I did e^y = x+1/x. Taking the limit x-> infinity for both sides, I got e^y -> 1, so y -> ln 1 = 0.
Therefore, lim an as n->infinity is 0 and an converges to 0.

I just realized that there's a completely easier way to prove divergence for this one. If the derivative of an diverges, then an should also diverge, since an has to be "larger" than its own derivative. Therefore,

an = ln (n+1) - ln n
an' = 1/(n+1) - 1/n

an' fails the p-series test, so both an' and an diverge.