# How to find the convergence/divergence of $\sum\limits_{n\geq 1} \frac{\sqrt {n-1}}{\sqrt {n(n+1)}}$

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.

• Thanks! I didn't know how to do the infinite symbol. – Alexander Ameye Jan 10 '18 at 21:22
• 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... – Shashi Jan 10 '18 at 22:16
• @AlexanderAmeye If you are ok, you can accept the answer and set as solved. Thanks! – user Jan 12 '18 at 23:41
• "General Term" $\displaystyle\sim {1 \over n^{\color{#f00}{1/2}}}$ as $\displaystyle n \to \infty$. – Felix Marin Jan 31 '18 at 23:54

$${{\sqrt{n-1}}\over{\sqrt{n(n+1)}}}\ge{{\sqrt{n-1}}\over{\sqrt{n^2+n+{1\over 4}}}}\ge{{\sqrt{n-1}}\over{{n+{1\over 2}}}}\ge{{1}\over{n+{1\over 2}}}$$

which is obviously divergent.

• Do you have any tips on quickly finding minorant sequences like these? – Alexander Ameye Jan 10 '18 at 21:26
• 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. – Mostafa Ayaz Jan 10 '18 at 21:29
• nice trick for the inequality! (+1) – user Jan 10 '18 at 21:45
• That's kind of you! – Mostafa Ayaz Jan 10 '18 at 21:50

Easier: for all $\;n>N\;$ , for some $\;N\in\Bbb N\;$ , we have that

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

• Ah right, so a minorant, divergent sequence! How can I quickly find a sequence like that? How did you come to $$\frac{\sqrt n}{\sqrt {2n^2}}$$ – Alexander Ameye Jan 10 '18 at 21:20
• @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. – zhw. Jan 10 '18 at 21:28
• @DonAntonio Sorry for my basic question but I really don't see that at the moment, how can justify the first inequality? Thanks! – user Jan 10 '18 at 21:31
• @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 – DonAntonio Jan 10 '18 at 21:42
• @DonAntonio ok thanks, thus it is not at first sight at least for mine :) – user Jan 10 '18 at 21:45

One more:

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

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

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.

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,

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