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I am trying to test the convergence of the series $$\sum_{n=2}^\infty n^p\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt n}\right)$$

You can find this series in exercise 8.15 (l) - Mathematical Analysis 2nd ed. - Apostol.

With some algebra I got $$\sum_{n=2}^\infty n^p\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt n}\right)=\sum_{n=2}^\infty\frac{n^p}{n\sqrt{n-1}+(n-1)\sqrt n}$$

I think I should use the comparison test. Hints on how to proceed?

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1 Answer 1

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$$\frac{n^p}{n\sqrt{n-1}+(n-1)\sqrt n}\sim\frac{n^p}{2n\sqrt n}=\frac12\cdot n^{p-3/2}$$

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  • $\begingroup$ What does that tilde stand for precisely? If $f(x)\sim g(x)$ then $\lim_{n\to \infty}\frac{f(x)}{g(x)}=1$? $\endgroup$
    – Charlie
    Sep 1, 2014 at 17:05
  • $\begingroup$ Could you explain me why those two quantities are "similar"? How did you choose $\frac{n^p}{2n\sqrt n}$? $\endgroup$
    – Charlie
    Sep 1, 2014 at 17:08
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    $\begingroup$ Yes, equivalent in the sense that the ratios converge to 1. Use $n-1\sim n$ and $\sqrt{n-1}\sim\sqrt{n}$. $\endgroup$
    – Did
    Sep 1, 2014 at 17:11

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