How do you test the alternating series $Sigma (-1)^(n+1)/sqrtn$ from n is $[1,oo)$ for convergence?

Question

17641 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-16T06:25:20

A sufficient condition for an alternating series $sum (-1)^na_n$ to converge is that:

(1) $lim a_n= 0$

(2) $a_(n+1) <= a_n$

The series:

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

is then convergent, since:

$lim 1/sqrt(n) = 0$

and

$1/sqrt(n+1) < 1/sqrt(n)$

How do you test the alternating series Sigma (-1)^(n+1)/sqrtnSigma (-1)^(n+1)/sqrtn from n is [1,oo)[1,oo) for convergence?