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- #1

- Thread starter calcboi
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- Thread starter
- #1

- Jan 17, 2013

- 1,667

Yes you are correct.

near infinity the term n^3 is dominant over ln (n).

near infinity the term n^3 is dominant over ln (n).

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- #3

- Jan 29, 2012

- 661

Right, $\displaystyle\lim_{n\to \infty}\left(\frac{1}{n^3-\log n}:\frac{1}{n^3}\right)=\ldots =1\neq 0$, so the series is absolutely convergent.BTW, I used the comparison c(x) = 1/n^3 for DCT and LCT since we know 1/n^3 converges by p-series.

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- #5

- Jan 29, 2012

- 661

The limit comparison test. Have a look hereWhat test did you use to determine absolute convergence? Or was that just analyzing end behavior?

Limit comparison test - Wikipedia, the free encyclopedia

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- #7

When I tried the Limit Comparison Test, I got infinity as n approaches infinity. How did you get 1?The limit comparison test. Have a look here

Limit comparison test - Wikipedia, the free encyclopedia

- Jan 29, 2012

- 661

$\displaystyle\lim_{n\to \infty}\left(\frac{1}{n^3-\log n}:\frac{1}{n^3}\right)=\lim_{n\to \infty}\frac{n^3}{n^3-\log n}=\lim_{n\to \infty}\frac{1}{1-(\log n/n^3)}=\frac{1}{1-0}=1$When I tried the Limit Comparison Test, I got infinity as n approaches infinity. How did you get 1?