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

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Answer 1 · 2016-12-27T14:04:37

$sum_(n=1)^oo (-1)^(n+1)/sqrt(n)$ is convergent but not absolutely convergent.

To determine if:

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

we can use use Leibniz' test, which states that a sufficient condition for an alternating series:

$sum_(n=1)^oo (-1)^(n)a_n$

to converge, is that the succession ${a_n}$ is decreasing and convergent to zero. In our case:

(i) $lim_n 1/sqrt(n) = 0$
(ii) $a_(n+1)/a_n = (1/sqrt(n+1))/(1/sqrt(n))=sqrt(n/(n+1)) < 1$

so the both conditions are met and the series is convergent.

We can also conclude that the series is not absolutely convergent, by direct comparison, since for any $n$ :

$1/sqrt(n) >= 1/n$

and

$sum_(n=1)^oo 1/n =oo$

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