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

##### 2 Answers

Alternating series, which alternate between having positive and negative terms, often come in the forms

Leibniz's rule, or the alternating series test, can be used to determine if one of these series converges or not.

For an alternating series such as **converge** if both:

#lim_(nrarroo)a_n=0" "" "# (the sequence approaches#0# )#a_n>=a_(n+1)" "" "# (the sequence is decreasing)

So, for

We see that

We can also show that

#1/(2n-1)>=1/(2(n+1)-1)=>1/(2n-1)>=1/(2n+1)#

And then by cross-multiplying and solving this inequality, which is tedious, we can show that this is always true (at least for

Anyway, we've showed that

Arranging and adding successive terms

so

and we have

so