How do you use the ratio test to test the convergence of the series #∑(x^(n))/(9^(n))# from n=1 to infinity?

The ratio test states the following:

1. Consider two consecutive terms #a_n# and #a_{n+1}#;
2. Divide the latter by the former and consider the absolute value: #abs(a_{n+1}/a_n)#;
3. Try to compute the limit of this ratio: #lim_{n\to\infty}abs(a_{n+1}/a_n)#;

THEN, if the limit exists:

• If it's bigger then #1# (strictly) , the series does not converge;
• If it's smaller then #1# (strictly), the series converges absolutely;
• If it equals one, the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.

In your case, #a_n=x^n/(9^n)#, and so #a_{n+1}=x^{n+1}/9^{n+1}#

Before dividing, it is useful to consider that:

• #x^{n+1}=x*x^n#,
• #9^{n+1}=9*9^n#
• dividing by a fraction means to multiply for the inverse of that fraction.

Now we can divide:

#a_{n+1}/a_n=(x*x^n)/(9*9^n) * 9^n/x^n#

Now the limit of #|x/9|# as #n\to infty# is #|x/9|# itself, because the quantity doesn't depend on #n#. So, the result of the test depends of #x#:

• If #|x|<9#, then #|x/9|<1# and the series is absolutely convergent;
• If #x=\pm 9# the limit is #1# and the test is inconclusive, but in this case it's easy to see that if #x=9# then you're summing infinite one's, and the sum is infinite, while if #x=-9# you're summing #(-1)^n#, i.e. #1-1+1-1+1-1+1...# and the series doesn't converge;
• If #|x|>9#, the series doesn't converge.

P.S., I know that the exercise mentioned explicitly the ratio test, but note that this case was much easier to solve noticing that

#sum x^n/9^n = sum (x/9)^n#, and thus converges iff #|x/9|<1#, confirming (of course!) what we just found.