Question

How do you determine if the series the converges conditionally, absolutely or diverges given `Sigma ((-1)^(n+1))/(n^1.5)` from `[1,oo)`?

Answer 1

The series:

`sum_(n=1)^oo (-1)^(n+1)/n^(3/2)`

is absolutely convergent

Explanation:

Given the series:

(1) `sum_(n=1)^oo (-1)^(n+1)/n^(3/2)`

we can test for convergence the series:

(2) `sum_(n=1)^oo abs((-1)^(n+1)/n^(3/2)) = sum_(n=1)^oo1/n^(3/2)`

If this series converges, then the series (1) converges absolutely ( and also conditionally ).

We can apply the integral test to the series (2) using the function:

`f(x) = 1/x^(3/2)`

As, for `x in [1,+oo)`, `f(x)` is positive and decreasing, it is infinitesimal for `x->+oo` and `f(n) = 1/n^(3/2)`.

Se we calculate:

`int_1^oo (dx)/x^(3/2) = [-2/x^(1/2)]_1^oo=2`

which means that the series (1) is absolutely convergent.