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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?

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2 Answers

up vote 2 down vote accepted

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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Suppose i use $\frac{3/2\pi}{n\sqrt{n}+n+1}$ as my upper bound, to prove that it is convergent i use comparison test with $\frac{3/2\pi}{n\sqrt{n}}$. Since $\frac{3/2\pi}{n\sqrt{n}}$ is convergent, then $\frac{3/2\pi}{n\sqrt{n}+n+1}$ is convergent. Therefore the inverse tan function series is convergent. Is my approach correct? –  uohzxela Oct 28 '12 at 3:02
    
@uohzxela: looks good to me. –  robjohn Oct 28 '12 at 7:48
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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 bound –  uohzxela Oct 28 '12 at 3:05
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