# How do you use the ratio test to test the convergence of the series #∑ (-5)^(n+1)n / 2^n# from n=1 to infinity?

See the explanation.

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To use the ratio test to test the convergence of the series ( \sum \frac{(-5)^{n+1}n}{2^n} ) from ( n = 1 ) to infinity, we compute the limit:

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

where ( a_n = \frac{(-5)^{n+1}n}{2^n} ).

Substitute ( a_{n+1} ) and ( a_n ) into the limit expression:

[ \lim_{n \to \infty} \left| \frac{\frac{(-5)^{(n+1)+1}(n+1)}{2^{(n+1)}}}{\frac{(-5)^{n+1}n}{2^n}} \right| ]

Simplify the expression:

[ \lim_{n \to \infty} \left| \frac{(-5)^{n+2}(n+1)2^n}{(-5)^{n+1}n2^{n+1}} \right| ]

[ = \lim_{n \to \infty} \left| \frac{(-5)(n+1)}{2(n)} \right| ]

[ = \lim_{n \to \infty} \left| \frac{-5n - 5}{2n} \right| ]

[ = \lim_{n \to \infty} \left| \frac{-5 - \frac{5}{n}}{2} \right| ]

[ = \frac{5}{2} ]

Since the limit is greater than 1, by the ratio test, the series ( \sum \frac{(-5)^{n+1}n}{2^n} ) diverges.

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