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

Question

2508 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-12T10:39:39

The series:

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

is absolutely convergent

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.