To test sum_(n=0)^oo r^n ∞∑n=0rn for convergence on the values rr
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
