To test #sum_(n=0)^oo r^n # for convergence on the values #r#
we can start by using the ratio test, where:
#lim_(n -> oo) abs(a_(n+1)/a_n) < 1# is convergent and divergent at #>1#
note it will be inconclusive at # =1#
so we get:
#lim_(n -> oo) abs(a_(n+1)/a_n) = lim_(n -> oo) abs(r^(n+1)/r^n)#
#= lim_(n -> oo) abs(r)#
#= abs(r)#
Thus we know that the series is convergent where # abs(r) < 1#
which mean, #-1 < r < 1# it is convergent.
then you would need to test the bounds (where #abs(r) = 1#) as the ratio test is inconclusive at that point.
so we test #sum_(n=0)^oo 1^n # for convergence, which is Divergent as if you write it out it would be #=1 + 1 + 1 + ... +1#
thus it does not converge to any finite number, and will sum to #oo#
Then we for #sum_(n=0)^oo (-1)^n # which is an alternating series,
and again we can see when writing the sum out we get #=1 - 1 + 1 - 1 + 1 .... -1 + 1#
thus we can see the sum would either equal #1# or #0# thus it does not converge a specific number, therefore we say it is Divergent
Thus we can say the series #sum_(n=0)^oo r^n # is Convergent where #r# is #RR (-1,1) #
or otherwise written # -1 < r < 1#
