Respuesta :
Calculating the limit as n goes to infinity, it is found that:
The sequence converges to 0.
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The sequence is given by:
[tex]a_n = \sum_{n=0}^{\infty} f(n)[/tex]
In which
[tex]f(n) = \frac{n^2 - 16}{n^3 + 4}[/tex]
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- If the limit of f(n) as n goes to infinity is a value, it converges to this value.
- If the limit of f(n) as n goes to infinity is infinity, it diverges.
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The limit is:
[tex]\lim_{n \rightarrow \infty} f(n) = \lim_{n \rightarrow \infty} \frac{n^2 - 16}{n^3 + 4}[/tex]
Since it is a limit going to infinity, we consider just the terms with the highest exponents in the numerator and the denominator. Thus:
[tex]\lim_{n \rightarrow \infty} \frac{n^2 - 16}{n^3 + 4} = \lim_{n \rightarrow \infty} \frac{n^2}{n^3} = \lim_{n \rightarrow \infty} \frac{1}{n} = \frac{1}{\infty} = 0[/tex]
Thus: The sequence converges to 0.
A similar question is given at https://brainly.com/question/20553793
