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?


1 Answer 1


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

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.