Question

How do you use the direct comparison test to determine if `Sigma 1/(n-1)` from `[1,oo)` is convergent or divergent?

Answer 1

Note that `1/(n-1)>1/n`.

You can think about this logically by comparing that `1/2>1/3`, `1/9>1/10`, and so on.

Since `sum_(n=2)^oo1/n` is divergent `1/(n-1)>1/n`, we see that `sum_(n=2)^oo1/(n-1)` is divergent as well through the direct comparison test.

(Note that the question asks for `sum_(n=1)^oo1/(n-1)`, which is undefined for `n=1`, so I've begun the series at `n=2` instead.)