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liberstina [14]
3 years ago
13

Use the ratio test to determine whether the series is convergent or divergent.2 +4/2^2 + 8/3^2 + 16/4^2 +

Mathematics
1 answer:
egoroff_w [7]3 years ago
3 0

Answer:

The result is 2 > 1. Hence, the series diverges

Step-by-step explanation:

You have the following series:

2+\frac{4}{2^2}+\frac{8}{3^2}+\frac{16}{4^2}+...+\frac{2^n}{n^2}=\sum_{i=1}^{i=n}a_n   (1)

You use the ratio test to determine if the series is convergent or divergent. The ratios test is given by:

\lim_{n \to \infty} \frac{a_{n+1}}{a_n} (2)

Then, you replace the series (1) in (2):

\lim_{n \to \infty} \frac{\frac{2^{n+1}}{(n+1)^2}}{\frac{2^n}{n^2}}= \lim_{n \to \infty} \frac{(2^n2)(n^2)}{2^n(n+1)^2}\\\\\lim_{n \to \infty} \frac{2n^2}{n^2+2n+1}= \lim_{n \to \infty} \frac{2n^2}{n^2[1+\frac{2}{n}+\frac{1}{n^2}]} \\\\\lim_{n \to \infty} \frac{2}{1+\frac{2}{n}+\frac{1}{n^2}}=\frac{2}{1+0+0}=2

In the ratio test you have that if the limit is greater than 1, the series diverges.

The result is 2 > 1. Hence, the series diverges.

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