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[SOLVED] Convergence 2

dwsmith

Well-known member
Feb 1, 2012
1,673
$\sum\limits_{n = 2}^{\infty}n^p\left(\frac{1}{\sqrt{n - 1}} - \frac{1}{\sqrt{n}}\right)$
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

Well-known member
Jan 26, 2012
890
$\sum\limits_{n = 2}^{\infty}n^p\left(\frac{1}{\sqrt{n - 1}} - \frac{1}{\sqrt{n}}\right)$
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
 

dwsmith

Well-known member
Feb 1, 2012
1,673

chisigma

Well-known member
Feb 13, 2012
1,704
$\displaystyle \frac{1}{\sqrt{n-1}}- \frac{1}{\sqrt{n}}= \frac{\sqrt{n}- \sqrt{n-1}} {\sqrt{n^{2}-n}}= \frac{1}{\sqrt{n^{2}-n}\ (\sqrt{n}+\sqrt{n+1})} = \frac{1}{\sqrt{n^{3}-n^{2}} + \sqrt{n^{3}-2\ n^{2} + n}} \sim \frac{1}{2} n^{-\frac{3}{2}}$

Kind regards

$\chi$ $\sigma$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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