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

Use the ratio test to find that this series diverges...

Then:

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To use the ratio test for the convergence of the series (\sum \frac{4^n}{3^n + 1}), we compute the limit:

[ L = \lim_{{n \to \infty}} \left|\frac{a_{n+1}}{a_n}\right| ]

where (a_n = \frac{4^n}{3^n + 1}).

First, find (a_{n+1}) and (a_n):

[ a_{n+1} = \frac{4^{n+1}}{3^{n+1} + 1}, \quad a_n = \frac{4^n}{3^n + 1} ]

Now, compute the ratio:

[ \frac{a_{n+1}}{a_n} = \frac{\frac{4^{n+1}}{3^{n+1} + 1}}{\frac{4^n}{3^n + 1}} ]

Simplify:

[ \frac{a_{n+1}}{a_n} = \frac{4^{n+1}(3^n + 1)}{4^n(3^{n+1} + 1)} = \frac{4(3^n + 1)}{3(3^{n+1} + 1)} ]

Simplify further:

[ \frac{a_{n+1}}{a_n} = \frac{4}{3} \cdot \frac{3^n + 1}{3^{n+1} + 1} ]

As (n) approaches infinity, (\frac{3^n + 1}{3^{n+1} + 1}) approaches (\frac{1}{3}). Therefore,

[ L = \frac{4}{3} \cdot \frac{1}{3} = \frac{4}{9} ]

Since (L < 1), by the ratio test, the series (\sum \frac{4^n}{3^n + 1}) converges.

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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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