Here is it: $\displaystyle \sum\limits_{n=1}^\infty \frac{ \sqrt{n+1} - \sqrt{n} }{n}$.
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Originally Posted by TWiX Here is it: $\displaystyle \sum\limits_{n=1}^\infty \frac{ \sqrt{n+1} - \sqrt{n} }{n}$. Use the LCT with the convergent $\displaystyle \frac{1}{n^{3/2}}$
Your series will coverge if $\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|<1$
Originally Posted by VonNemo19 Use the LCT with the convergent $\displaystyle \frac{1}{n^{3/2}}$ Failed. The limit $\displaystyle =\infty$
Originally Posted by pickslides Your series will coverge if $\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|<1$ I didnt try it. But its algebric function. I think it will be 1 ---> Failed.
Originally Posted by pickslides Your series will coverge if $\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|<1$ That's hard to show, no?
$\displaystyle \frac{\sqrt{n+1}-\sqrt{n}}{n}=\frac{1}{n\left( \sqrt{n+1}+\sqrt{n} \right)}<\frac{1}{2n^{\frac{3}{2}}}.$
In general not that bad, but this question is a bit sticky yep
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