To determine whether the series
[tex]\[
\sum_{n=1}^{\infty} \frac{n-4}{\sqrt{n^3+n^2+8}}
\][/tex]
converges or diverges, we can use the Limit Comparison Test. The Limit Comparison Test is particularly useful when dealing with series whose terms are complicated but bear a resemblance to terms of a simpler series that we know how to handle.
Let's start by considering a simpler comparison series. Given the [tex]\( n^3 \)[/tex] term in the denominator, which dominates as [tex]\( n \)[/tex] becomes very large, a reasonable comparison might be the series
[tex]\[
\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}.
\][/tex]
We know that the series [tex]\(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\)[/tex] diverges (this is a p-series with [tex]\( p = \frac{1}{2} \)[/tex], and it diverges because [tex]\( \frac{1}{2} \le 1 \)[/tex]).
To apply the Limit Comparison Test, we need to evaluate the limit:
[tex]\[
\lim_{n \to \infty} \frac{\frac{n-4}{\sqrt{n^3+n^2+8}}}{\frac{1}{\sqrt{n}}}
\][/tex]
Simplify the expression inside the limit:
[tex]\[
\lim_{n \to \infty} \frac{n-4}{\sqrt{n^3+n^2+8}} \cdot \sqrt{n}
\][/tex]
[tex]\[
= \lim_{n \to \infty} \frac{(n-4)\sqrt{n}}{\sqrt{n^3+n^2+8}}
\][/tex]
Notice that the dominant term in the numerator is [tex]\( n \sqrt{n} = n^{3/2} \)[/tex], and the dominant term in the denominator is [tex]\( \sqrt{n^3} = n^{3/2} \)[/tex]:
[tex]\[
= \lim_{n \to \infty} \frac{n^{3/2} - 4\sqrt{n}}{n^{3/2}\sqrt{1 + \frac{1}{n} + \frac{8}{n^3}}}
\][/tex]
[tex]\[
= \lim_{n \to \infty} \frac{n^{3/2} - 4\sqrt{n}}{n^{3/2}}\sqrt{\frac{1}{1 + \frac{1}{n} + \frac{8}{n^3}}}
\][/tex]
Since [tex]\( \frac{1}{n} \)[/tex] and [tex]\( \frac{8}{n^3} \)[/tex] both tend to 0 as [tex]\( n \to \infty \)[/tex], we have:
[tex]\[
\sqrt{1 + \frac{1}{n} + \frac{8}{n^3}} \to \sqrt{1} = 1
\][/tex]
Thus,
[tex]\[
= \lim_{n \to \infty} \left(1 - \frac{4}{n^{3/2}}\right) = 1
\][/tex]
The limit we evaluated is 1, which is a positive finite number. According to the Limit Comparison Test, since
[tex]\[
\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}
\][/tex]
diverges and our limit is positive and finite, the original series
[tex]\[
\sum_{n=1}^{\infty} \frac{n-4}{\sqrt{n^3+n^2+8}}
\][/tex]
also diverges.
Thus, the series diverges.
