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?

Question

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?

Christopher Agresta · Accepted Answer

See the explanation.

Paisley Johnson · Answer

undefined 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.