- #1

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A.) for n > 2, [tex]\frac{n}{n^3-4}[/tex] < [tex]\frac{2}{n^2}[/tex], and the series [tex]2\sum \frac{1}{n^2}[/tex] converges, so by the comparison test, the series [tex]\sum \frac{n}{n^3-4}[/tex] converges

B.) for n > 2, [tex]\frac{ln(n)}{n^2}[/tex] > [tex]\frac{1}{n^2}[/tex], and the series [tex]\sum \frac{1}{n^2}[/tex] converges, so by the comparison test, the series [tex]\sum \frac{ln(n)}{n^2}[/tex] converges

C.) for n > 2, [tex]\frac{1}{n^2-6}[/tex] < [tex]\frac{1}{n^2}[/tex], and the series [tex]\sum \frac{1}{n^2}[/tex] converges, so by the comparison test, the series [tex]\sum \frac{1}{n^2-6}[/tex] converges

For A.) and C.) sounds true, by the p-series, both converges so that means mean the original series converges, so A.) and C.) are True right?

and B.) I also think it's true by the p-series.