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How do I find the convergence of the following sequence? $$\sum_{n\geq 1}\frac{\sqrt {n-1}}{\sqrt {n(n+1)}}$$

I have tried both the root test (Cauchy) and the ratio test (d'Alembert).

They both were inconclusive.

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  • $\begingroup$ Thanks! I didn't know how to do the infinite symbol. $\endgroup$ – Alexander Ameye Jan 10 '18 at 21:22
  • $\begingroup$ You're welcome. One more thing: you wanted to start the summation from $n=1$, right? As it is now your first term is not defined... $\endgroup$ – Shashi Jan 10 '18 at 22:16
  • $\begingroup$ @AlexanderAmeye If you are ok, you can accept the answer and set as solved. Thanks! $\endgroup$ – user Jan 12 '18 at 23:41
  • $\begingroup$ "General Term" $\displaystyle\sim {1 \over n^{\color{#f00}{1/2}}}$ as $\displaystyle n \to \infty$. $\endgroup$ – Felix Marin Jan 31 '18 at 23:54
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$${{\sqrt{n-1}}\over{\sqrt{n(n+1)}}}\ge{{\sqrt{n-1}}\over{\sqrt{n^2+n+{1\over 4}}}}\ge{{\sqrt{n-1}}\over{{n+{1\over 2}}}}\ge{{1}\over{n+{1\over 2}}}$$

which is obviously divergent.

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  • $\begingroup$ Do you have any tips on quickly finding minorant sequences like these? $\endgroup$ – Alexander Ameye Jan 10 '18 at 21:26
  • $\begingroup$ Well! when tending $n\to \infty$ the sequence behavior is very like that of ${1\over{n+{1\over 2}}}$ meanwhile $a_n\sim{1\over{n+{1\over 2}}}$ for sufficiently large n. So must be the convergence of $a_n$ like that of its limit when in $\infty$. This gives you an intuition and you need to make it more precise for a formal proof. $\endgroup$ – Mostafa Ayaz Jan 10 '18 at 21:29
  • $\begingroup$ nice trick for the inequality! (+1) $\endgroup$ – user Jan 10 '18 at 21:45
  • $\begingroup$ That's kind of you! $\endgroup$ – Mostafa Ayaz Jan 10 '18 at 21:50
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Easier: for all $\;n>N\;$ , for some $\;N\in\Bbb N\;$ , we have that

$$\frac{\sqrt{n-1}}{\sqrt{n(n+1)}}\ge\frac{\sqrt n}{\sqrt{2n^2}}=\frac{\sqrt n}{\sqrt2\,n}=\frac1{\sqrt2\,\sqrt n}$$

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  • $\begingroup$ Ah right, so a minorant, divergent sequence! How can I quickly find a sequence like that? How did you come to $$ \frac{\sqrt n}{\sqrt {2n^2}}$$ $\endgroup$ – Alexander Ameye Jan 10 '18 at 21:20
  • $\begingroup$ @AlexanderAmeye You ask yourself what the "size" is of the top and bottom. $\sqrt {n-1}$ is like $\sqrt n,$ and $\sqrt {n(n+1)}$ is like $\sqrt {n^2} = n.$ Thus the quotient is like $1/\sqrt n.$ That's not a proof, but your question was how to come to it. Having a gut feeling this way at the outset is a valuable beginning. $\endgroup$ – zhw. Jan 10 '18 at 21:28
  • $\begingroup$ @DonAntonio Sorry for my basic question but I really don't see that at the moment, how can justify the first inequality? Thanks! $\endgroup$ – user Jan 10 '18 at 21:31
  • $\begingroup$ @gimusi, there are no basic questions. Just (good, bad or medium) question...:) : $$\frac{\sqrt{n-1}}{\sqrt{n(n+1)}}\ge\frac{\sqrt n}{\sqrt{2n^2}}=\frac{\sqrt n}{\sqrt2\,n}\iff\sqrt2\,n\sqrt{n-1}\ge n\sqrt{n+1}\iff2\ge1+\frac2{n-1}$$ and the last inequality is easy to check for all $\;n\;$ big enough $\endgroup$ – DonAntonio Jan 10 '18 at 21:42
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    $\begingroup$ @DonAntonio ok thanks, thus it is not at first sight at least for mine :) $\endgroup$ – user Jan 10 '18 at 21:45
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One more:

$\dfrac{\sqrt{n-1}}{\sqrt{n(n+1)}} \gt \dfrac{\sqrt{n/2}}{(n+1)}\gt $

$\dfrac{\sqrt{n/2}}{2n} \gt \dfrac{1}{4\sqrt{n}}.$

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You can prove that for any $n\geq 2$ the inequality $$ \frac{\sqrt{n-1}}{\sqrt{n(n+1)}}\geq 2\sqrt{n+5}-2\sqrt{n+4} $$ holds, hence $$ \sum_{n=1}^{N}\frac{\sqrt{n-1}}{\sqrt{n(n+1)}}=\sum_{n=2}^{N}\frac{\sqrt{n-1}}{\sqrt{n(n+1)}}\geq 2\sum_{n=2}^{N}\left(\sqrt{n+5}-\sqrt{n+4}\right)=2\sqrt{N+5}-2\sqrt{6} $$ and the original series is clearly divergent. The main advantage of this approach (creative telescoping) is that you also get an accurate estimation on the growth of the partial sums.

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since, $\frac{1}{n}\ge\frac{1}{n+1}$ and for $n\ge 2$ we have $$\sqrt{1-\frac{ 2}{n+1}}\ge \frac14$$ hence,

$$\frac{\sqrt{n-1}}{\sqrt{n(n+1)}}\ge\frac{\sqrt {n-1}}{n+1}=\frac{1}{\sqrt{n+1}}\sqrt{\frac{ n-1}{n+1}}=\frac{1}{\sqrt{n+1}}\sqrt{1-\frac{ 2}{n+1}}\ge \frac{1}{2\sqrt{n+1}}$$

Therefore,

$$\sum \frac{\sqrt{n-1}}{\sqrt{n(n+1)}}\ge \sum\frac{1}{2\sqrt{n+1}} =\infty.$$

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