How do you determine if the series the converges conditionally, absolutely or diverges given #Sigma ((-1)^(n))/(sqrt(n+4))# from #[1,oo)#?

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Answer 1 · 2017-08-08T23:03:24

The series:

#sum_(n=1)^oo (-1)^n/sqrt(n+4) #

is convergent but not absolutely convergent.

This is an alternating series in the form:

#sum_(n=1)^oo (-1)^na_n#,

thus based on Leibniz's theorem, it is convergent if:

#(1) lim_(n->oo) a_n = 0#

#(2) a_n >= a_(n+1)#

In our case:

#lim_(n->oo) 1/sqrt(n+4) = 0#

and

#1/sqrt(n+4) > 1/sqrt(n+5)#

so both conditions are satisfied and the series is convergent.

Consider now the series of absolute values:

#(3) sum_(n=1)^oo 1/sqrt(n+4)#

Using the limit comparison test we can see that as:

#lim_(n->oo) sqrt(n)/sqrt(n+4) = 1#

the series #(3)# has the same character as the series:

#sum_(n=1)^oo 1/sqrt(n) = sum_(n=1)^oo 1/n^(1/2)#

which is divergent based on the #p#-series test.

The series #(3)# is then also divergent, so the series:

#sum_(n=1)^oo (-1)^n/sqrt(n+4) #

is convergent but not absolutely convergent.