Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The series $$\sum_{n=1}^{+\infty} \frac{1}{\sqrt{n(n+1)}}$$ I have tried ratio and integral both lead me to inconclusive, so probably it's by comparisson but I can't find What to compare.

share|improve this question
4  
The integral test is inconclusive? What did you get for $$\int_1^{\infty}{dx\over\sqrt{x(x+1)}}$$ –  Gerry Myerson Aug 13 at 13:22

1 Answer 1

Try a Limit Comparison with $\sum \frac{1}{n}$.

Along similar lines, note that $n(n+1)\lt n(4n)$, so $\dfrac{1}{\sqrt{n(n+1)}}\gt \dfrac{1}{2n}$.

The informal idea is that for large $n$, the number $\sqrt{n(n+1)}$ is roughly equal to $n$.

share|improve this answer
    
I get n/(sqrt(n(n+1))), I should simplifie more? –  user169504 Aug 13 at 13:27
    
Yes, for the limit comparison you then divide top and bottom by $n$. You will get $\frac{1}{\sqrt{1+\frac{1}{n}}}$, and the limit is $1$. My second suggestion is more "concrete" so you may prefer it. –  André Nicolas Aug 13 at 13:34
    
So for the limit comparison the algebraic "trick" one uses is $\sqrt{n(n+1)}=n\sqrt{1+\frac{1}{n}}$. –  André Nicolas Aug 13 at 13:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.