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

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Answer 1 · 2017-01-30T20:57:42

the series is indeterminate.

We can easily see that the series is not convergent, since:

$lim_(n->oo) (-1)^n != 0$

We can take a closer look at the partial sums:

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

$s_1 = -1$
$s_2 =0$
$s_3 = -1$
$...$

and we can prove by induction that:

${(s_(2n) = 0),(s_(2n+1) = -1):}$

so that partial sums oscillate between the two values and do not converge to a limit.

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