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Timed Test For
Direct comparison test
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Sample Problem
We know that $\displaystyle \dfrac{n^2}{n^3-\ln(n)} \ge \dfrac{n^2}{n^3}=\dfrac{1}{n}>0$ for any $n\ge 1$. **Considering this fact, what does the direct comparison test say about $\displaystyle\sum\limits_{n=1}^{\infty }~\dfrac{n^2}{n^3-\ln(n)}$ ?** [[☃ radio 3]]
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Duration
30 seconds
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