Question

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

Answer 1

The series converges if `x \in (-9,9)`.

Explanation:

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.