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$ \sum_{n=1}^{\infty}\cfrac{\pi + \tan^{-1}n}{n\sqrt{n}+n+1}$

Hi all, these are questions from my graded math homework. For the first qn, I don't know how to proceed, coz of the inverse tan function. I'm clueless on whether i should use Comparison Test or Limit Comparison Test. Any ideas?

$ \sum_{n=1}^{\infty}\frac{\sqrt[4]{2n^8-4n^4+n}}{\sqrt[3]{n^7-3n^5+n^3}}$

For the second question, i am using Limit Comparison Test with $\cfrac{\sqrt[4]{2n^8}}{\sqrt[3]{n^7}}$ as the denominator, and the limit is 1. Since 1 is a positive real number, I can deduce that the series is divergent as $\cfrac{\sqrt[4]{2n^8}}{\sqrt[3]{n^7}}$ is divergent. Is this correct?

2 Answers 2

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Hint: $-\frac\pi2\lt\tan^{-1}(x)\lt\frac\pi2$

Hint: $\displaystyle \frac{\sqrt[4]{2n^8-4n^4+n}}{\sqrt[3]{n^7-3n^5+n^3}}=n^{-1/3}\frac{\sqrt[4]{2-4n^{-4}+n^{-7}}}{\sqrt[3]{1-3n^{-2}+n^{-4}}}$

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    @uohzxela: looks good to me.2012-10-28
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HINT for (1): Note that $\tan^{-1}x$ is bounded. (By what?)

Your approach to (2) is fine, though it isn’t correct to say that $\cfrac{\sqrt[4]{2n^8}}{\sqrt[3]{n^7}}$ is divergent. What you meant is that $\sum_{n\ge 1}\cfrac{\sqrt[4]{2n^8}}{\sqrt[3]{n^7}}$ is divergent, which is true.

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    Thanks for pointing that out regarding the n >= 1 bound2012-10-28