Math  /  Calculus

QuestionConsider the following series. n=13n214n5+3n+1\sum_{n=1}^{\infty} \frac{3 n^{2}-1}{4 n^{5}+3 n+1}
Use the Limit Comparison Test to complete the limit.

Studdy Solution
Interpret the result of the limit. Since the limit is a positive finite number (1), the Limit Comparison Test tells us that the original series n=13n214n5+3n+1\sum_{n=1}^{\infty} \frac{3n^2-1}{4n^5+3n+1} converges or diverges together with the comparison series n=134n3\sum_{n=1}^{\infty} \frac{3}{4n^3}.
Since n=11n3\sum_{n=1}^{\infty} \frac{1}{n^3} is a convergent pp-series with p=3>1p = 3 > 1, the original series also converges.
The series n=13n214n5+3n+1\sum_{n=1}^{\infty} \frac{3n^2-1}{4n^5+3n+1} converges.

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