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

By the ratio test, the series converges.

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To use the ratio test to test the convergence of the series ( \sum_{n=1}^{\infty} \frac{n+1}{3^n} ), you would calculate the limit:

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

Where ( a_n ) represents the nth term of the series. In this case, ( a_n = \frac{n+1}{3^n} ).

So, we have:

[ \lim_{n \to \infty} \left| \frac{\frac{n+2}{3^{n+1}}}{\frac{n+1}{3^n}} \right| ]

[ = \lim_{n \to \infty} \left| \frac{n+2}{3^{n+1}} \times \frac{3^n}{n+1} \right| ]

[ = \lim_{n \to \infty} \left| \frac{n+2}{3n + 3} \right| ]

Since the numerator and denominator both approach infinity as ( n ) approaches infinity, we can apply L'Hopital's Rule to evaluate the limit:

[ = \lim_{n \to \infty} \frac{1}{3} = \frac{1}{3} ]

According to the ratio test, if this limit is less than 1, the series converges absolutely. If it's greater than 1 or diverges to infinity, the series diverges. If the limit equals 1, the ratio test is inconclusive, and another test may be needed. In this case, since ( \frac{1}{3} ) is less than 1, the series ( \sum_{n=1}^{\infty} \frac{n+1}{3^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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