Question

How do you determine the convergence or divergence of `Sigma ((-1)^(n))/(sqrtn)` from `[1,oo)`?

Answer 1

The sum is convergent.

Explanation:

`sum_(k=1)^oo(-1)^k/sqrt(k)`

is an alternate series with `abs(a_k) < abs(a_(k-1))` so it converges.

Here `a_k = 1/sqrt(k)`.