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)#.

The ratio test states the following:

THEN, if the limit exists:

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:

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#:

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.
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Answer 2

To use the ratio test for the series ∑(x^(n))/(9^(n)), calculate the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term. In this case, it would be lim(n→∞) |(x^(n+1))/(9^(n+1))| / |(x^(n))/(9^(n))|. Simplify this expression to |(x^(n+1))/9^(n+1)| * |9^n / x^n|. This simplifies to |x / 9| as n approaches infinity. If |x / 9| is less than 1, the series converges. If |x / 9| is greater than 1, the series diverges. If |x / 9| equals 1, the test is inconclusive, and another test may be needed.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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