# How do you use the ratio test to test the convergence of the series #∑ (3/4)^n# from n=1 to infinity?

The series is convergent and:

The ratio to test is:

We can note that this is a particular case of the geometric series:

So that we can also calculate the sum:

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To use the ratio test to test the convergence of the series (\sum_{n=1}^\infty \left(\frac{3}{4}\right)^n), you need to compute the limit of the ratio of consecutive terms as (n) approaches infinity.

The general form of the ratio test states: If (L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|), then:

- If (L < 1), the series converges absolutely.
- If (L > 1), the series diverges.
- If (L = 1), the test is inconclusive.

For the given series, (a_n = \left(\frac{3}{4}\right)^n), so (a_{n+1} = \left(\frac{3}{4}\right)^{n+1}).

Now, compute the limit: [L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\left(\frac{3}{4}\right)^{n+1}}{\left(\frac{3}{4}\right)^n} \right|] [= \lim_{n \to \infty} \left| \frac{\left(\frac{3}{4}\right)^{n}\cdot\frac{3}{4}}{\left(\frac{3}{4}\right)^n} \right|] [= \lim_{n \to \infty} \left| \frac{\frac{3}{4}}{1} \right|] [= \frac{3}{4}]

Since (L = \frac{3}{4} < 1), by the ratio test, the series (\sum_{n=1}^\infty \left(\frac{3}{4}\right)^n) converges absolutely.

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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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