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.

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