Sum of an Infinite Series with Real Exponent p

[Dustinsfl]

$\sum\limits_{n = 2}^{\infty}n^p\left(\frac{1}{\sqrt{n - 1}} - \frac{1}{\sqrt{n}}\right)$
where p is any fixed real number.
If this was just the telescoping series or the p-series, this wouldn't be a problem.

 

[CaptainBlack]

$\sum\limits_{n = 2}^{\infty}n^p\left(\frac{1}{\sqrt{n - 1}} - \frac{1}{\sqrt{n}}\right)$
where p is any fixed real number.
If this was just the telescoping series or the p-series, this wouldn't be a problem.

\[ \left( \frac{1}{\sqrt{n - 1}} - \frac{1}{\sqrt{n}} \right) \sim \frac{n^{-3/2}}{2}\]

CB

 

[Dustinsfl]

\[ \left( \frac{1}{\sqrt{n - 1}} - \frac{1}{\sqrt{n}} \right) \sim \frac{n^{-3/2}}{2}\]

CB

How did you come up with that though?

 

[chisigma]

$\displaystyle \frac{1}{\sqrt{n-1}}- \frac{1}{\sqrt{n}}= \frac{\sqrt{n}- \sqrt{n-1}} {\sqrt{n^{2}-n}}= \frac{1}{\sqrt{n^{2}-n}\ (\sqrt{n}+\sqrt{n+1})} = \frac{1}{\sqrt{n^{3}-n^{2}} + \sqrt{n^{3}-2\ n^{2} + n}} \sim \frac{1}{2} n^{-\frac{3}{2}}$

Kind regards

$\chi$ $\sigma$

 

[CaptainBlack]

How did you come up with that though?

\[\begin{aligned}\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}&=\frac{1}{\sqrt{n}\sqrt{1-1/n}}-\frac{1}{\sqrt{n}}\\&= \frac{1}{\sqrt{n}}\left(1+(-1/2)(-n)^{-1}+O(n^{-2})\right)-\frac{1}{\sqrt{n}}\\ & = \dots \end{aligned}\]

CB