MHB Sum of an Infinite Series with Real Exponent p

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The discussion centers on the convergence of the infinite series $\sum_{n=2}^{\infty} n^p \left(\frac{1}{\sqrt{n-1}} - \frac{1}{\sqrt{n}}\right)$ for any fixed real exponent p. Participants analyze the behavior of the term $\frac{1}{\sqrt{n-1}} - \frac{1}{\sqrt{n}}$, noting that it approximates to $\frac{n^{-3/2}}{2}$. This approximation is derived through algebraic manipulation and limits, demonstrating how the series can be simplified. The convergence of the series is influenced by the growth rate of the terms involved, particularly as n increases. Ultimately, the discussion highlights the mathematical intricacies of evaluating such infinite series.
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$\sum\limits_{n = 2}^{\infty}n^p\left(\frac{1}{\sqrt{n - 1}} - \frac{1}{\sqrt{n}}\right)$
where p is any fixed real number.
If this was just the telescoping series or the p-series, this wouldn't be a problem.
 
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dwsmith said:
$\sum\limits_{n = 2}^{\infty}n^p\left(\frac{1}{\sqrt{n - 1}} - \frac{1}{\sqrt{n}}\right)$
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
 
CaptainBlack said:
\[ \left( \frac{1}{\sqrt{n - 1}} - \frac{1}{\sqrt{n}} \right) \sim \frac{n^{-3/2}}{2}\]

CB

How did you come up with that though?
 
$\displaystyle \frac{1}{\sqrt{n-1}}- \frac{1}{\sqrt{n}}= \frac{\sqrt{n}- \sqrt{n-1}} {\sqrt{n^{2}-n}}= \frac{1}{\sqrt{n^{2}-n}\ (\sqrt{n}+\sqrt{n+1})} = \frac{1}{\sqrt{n^{3}-n^{2}} + \sqrt{n^{3}-2\ n^{2} + n}} \sim \frac{1}{2} n^{-\frac{3}{2}}$

Kind regards

$\chi$ $\sigma$
 
dwsmith said:
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
 
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