Respuesta :

Answer:

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

Step-by-step explanation:

You have the following series:

[tex]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[/tex]   (1)

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

[tex]\lim_{n \to \infty} \frac{a_{n+1}}{a_n}[/tex] (2)

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

[tex]\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[/tex]

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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