How do you determine the convergence or divergence of $Sigma ((-1)^(n+1)n)/(2n-1)$ from $[1,oo)$ ?

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Answer 1 · 2016-12-23T09:28:23

The series:
$sum_(n=1)^oo (-1)^(n+1)n/(2n-1)$
is divergent.

You can determine whether an alternating series converges using Leibniz' criteria, which states that:

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

converges if:

(i) $a_n>a_(n+1)$
(ii) $lim_n a_n = 0$

As the general term of the series above can be expressed as:

$a_n = -n/(2n-1)$

We can quickly see that:

$lim_n a_n = lim_n-n/(2n-1)= lim_n-1/(2-1/n)=-1/2$

so that condition (ii) is not met and the series is divergent.

How do you determine the convergence or divergence of Sigma ((-1)^(n+1)n)/(2n-1)Sigma ((-1)^(n+1)n)/(2n-1) from [1,oo)[1,oo)?