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

By ratio test I get $$ \lim_{n\to \infty} \frac{\frac{\sqrt{n(n+1)}-n}{n\sqrt{n(n+1)}}}{\frac{1}{n}}=\frac{0}{0}$$

So i can't conclude anything. I tried comparison tests, but i get $\le$ something divergent.

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  • $\begingroup$ Surely you could do a better attempt at the limit? Flip those fractions and L'hospital's rule like pancakes on a stove. $\endgroup$ – Simply Beautiful Art Sep 2 '16 at 12:05
  • $\begingroup$ Look in your book for the material on "indeterminate forms". $\endgroup$ – GEdgar Sep 2 '16 at 12:11
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\begin{align*} \sum_{n=1}^{\infty}\frac{\sqrt{n(n+1)}-n}{n\sqrt{n(n+1)}} &=\sum_{n=1}^{\infty}\frac{(\sqrt{n(n+1)}-n)(\sqrt{n(n+1)}+n)}{n\sqrt{n(n+1)}(\sqrt{n(n+1)}+n)}\\ &=\sum_{n=1}^{\infty}\frac1{\sqrt{n(n+1)}(\sqrt{n(n+1)}+n)}\\ &=\sum_{n=1}^{\infty}\frac1{n(n+1)+n\sqrt{n(n+1)}}\\ &\le\sum_{n=1}^{\infty}\frac1{n^2}. \end{align*}

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  • $\begingroup$ Brilliant. thanks $\endgroup$ – Rush ThaMan Sep 2 '16 at 12:12
  • $\begingroup$ @RushThaMan You're welcome! $\endgroup$ – Cm7F7Bb Sep 2 '16 at 12:17
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Use Taylor's expansion at order $1$: \begin{align*}\frac1n-\frac1{\sqrt{n(n+1)}}&=\frac1n\Biggl(1-\frac1{\sqrt{1+\frac1n}}\Biggr)=\frac1n\biggl(1-1+\frac1{2n}+o\Bigl(\frac1n\Bigr)\biggr)\\ &=\frac1{2n^2}+o\biggl(\frac1{n^2}\biggr)\sim_\infty\frac1{2n^2} \end{align*} and the latter converges.

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Compare to $$\sum_{n=1}^\infty\left(\frac1n-\frac1{\sqrt{n\cdot n}}\right)$$ on one side and $$\sum_{n=1}^\infty\left(\frac1n-\frac1{\sqrt{(n+1)(n+1)}}\right)$$on the other.

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You don't need ratio test or a clever bound for the general term... Like most problems on MSE, a simple asymptotic expansion fits the bill.

$$\frac{1}{n}-\frac{1}{\sqrt{n(n+1)}} = \frac{1}{2n^2} + o\left(\frac{1}{n^2}\right)$$

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