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We have the following series:

$$\sum_{n=1}^{\infty} \frac{\sqrt{n^3+n}-\sqrt{n^3}}{\sqrt{n(n+1)}-1}$$

According to WolframAlpha it is convergent but I don't see why. Intuitively, the expression under the sum must be assymptotically similair to something like $\frac{1}{n^\alpha}$, and clearly $\sqrt{n^2}$ dominates the denominator. In the numerator the expression looks similair either to $\sqrt{n^3}$ but $\frac{\sqrt{n^3}}{\sqrt{n^2}}={\sqrt{n}}$ which gives as assymptotic similarity to sequence, the sum of which is divergent. On the other hand, if $\sqrt{n}$ dominates the numerator we get $\frac{\sqrt{n}}{\sqrt{n^2}}=\frac{1}{\sqrt{n}}$ so again assymptotic similarity to sequence, which gives divergent sum.

So why is this sum convergent anyway?

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    $\begingroup$ rationalize the numerator $\endgroup$ – Will Jagy Dec 7 '14 at 19:04
  • $\begingroup$ We get $\frac{n}{\sqrt{(n+1)n(n^3+n)}+\sqrt{n^4(n+1)}-\sqrt{n^3+n}-\sqrt{n^3}}$. Is this assymptotically similair to $\frac{1}{n^{\frac{3}{2}}}$? $\endgroup$ – qiubit Dec 7 '14 at 19:10
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You may write, as $n$ is great: $$ \begin{align} \frac{\sqrt{n^3+n}-\sqrt{n^3}}{\sqrt{n(n+1)}-1}&=\frac{(\sqrt{n^3+n}-\sqrt{n^3})(\sqrt{n(n+1)}+1)}{n(n+1)-1}\\\\ &=\frac{n^{5/2} }{n^2}\frac{\left(\sqrt{1+\frac{1}{n^2}}-1\right)\left(\sqrt{1+\frac{1}{n^2}}+\frac1n\right)}{1+\frac{1}{n}-\frac{1}{n^2}}\\\\ &=\frac{n^{5/2} }{n^2}\frac{\left(\frac{1}{n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)\left(1+\mathcal{O}\left(\frac{1}{n}\right)\right)}{1+\frac{1}{n}-\frac{1}{n^2}}\\\\ & \sim \frac{n^{5/2}}{n^4}=\frac{1}{n^{3/2} } \end{align} $$ and your series is convergent by the comparison test.

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We have

$$\frac{\sqrt{n^3+n}-\sqrt{n^3}}{\sqrt{n(n+1)}-1}\sim_\infty\frac{n^{3/2}((1+n^{-2})^{1/2}-1)}{n}\sim_\infty\frac1{2n^{3/2}}$$ and the Riemann series $$\sum \frac1{n^{3/2}}$$ is convergent so the given series is convergent by asymptotic comparison.

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