I don't understand this explanation for #\sum_(n=0)^\infty((-1)^n)/(5n-1)#? Why test for convergence/divergence AGAIN, if the Limit Comparison Test confirms that both series are the same?

The Limit Comparison Test, when applied to this problem, tells us that the absolute value of the series, #sum_(n=0)^oo|a_n|,# diverges.

However, it is totally possible to have #sum_(n=0)^oo|a_n|# diverge and #sum_(n=0)^ooa_n# converge.

This is called conditional convergence , and we must always check for it.

There is no "absolute divergence" that tells us if the absolute value of the series diverges, so must the series itself.

So, if #a_n=(-1)^n/(5n+1), |a_n|=1/(5n+1)# as the absolute value omits the alternating negative signs.

Now, we can say #b_n=1/(5n)#, and we know #sum_(n=0)^oo1/(5n)=1/5sum_(n=0)^oo1/n# diverges, it is a #p#-series with #p=1#.

Now, using the Limit Comparison Test,

#c=lim_(n->oo)|a_n|/b_n=(1/(5n+1))/(1/(5n))=lim_(n->oo)(5n)/(5n+1)=1>0neoo#

So, both series must diverge.

Now, the original series, #sum_(n=0)^oo(-1)^n/(5n+1)# is an alternating series -- we are not taking the absolute value; we are acknowledging that it has negative terms.

Now, the positive, non-alternating portion of #a_n=(-1)^n/(5n+1)# is #b_n=1/(5n+1)#

The Alternating Series Test tells us if

#lim_(n->oo)b_n=0, b_n>=b_(n+1)# (IE #b_n# is decreasing), then the series converges.

#1/(5n-1)# decreases as #n# grows due to the growing denominator; #lim_(n->oo)1/(5n-1)=0#, so the series converges.