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.

It's one of my homework exercises that is rather problematic to me. Apparently the last thing to do is to squeeze it but I don't see yet how to do that. Could you help?

$$\lim_{n\to\infty}\frac{\displaystyle\sum_{k=1}^{n}\frac{2n}{k^2+kn+n^2}-\int_0^1\frac{1}{x^2-x+1}dx}{\displaystyle\sum_{k=1}^{n}\frac{n}{k^2+kn+n^2}-\int_0^1\frac{1}{x^2+x+1}dx}$$

share|improve this question
    
I suppose there is a typo in the formula. –  Ma Ming May 5 '13 at 20:37
1  
Welcome to math.SE! I have tried to improve the readability of your question by improving TeX. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened. For some basic information about writing math at this site see e.g. here and here. –  Américo Tavares May 5 '13 at 20:37
    
@AméricoTavares thank you. The question is correct. –  Peter Wormy May 5 '13 at 20:41
    
You are welcome! –  Américo Tavares May 5 '13 at 20:42

1 Answer 1

We can prove easily by change of variable that $$2\int_0^1\frac{dx}{x^2+x+1}=\int_0^1\frac{dx}{x^2-x+1}=\frac{2\sqrt{3}\pi}{9}=2\alpha$$ so we have $$\lim_{n\to\infty}\frac{\displaystyle\sum_{k=1}^{n}\frac{2n}{k^2+kn+n^2}-\int_0^1\frac{1}{x^2-x+1}dx}{\displaystyle\sum_{k=1}^{n}\frac{n}{k^2+kn+n^2}-\int_0^1\frac{1}{x^2+x+1}dx}=\lim_{n\to\infty}\frac{2S_n-2\alpha}{S_n-\alpha}=2$$

Added

\begin{align}\int_0^1\frac{dx}{x^2+x+1}&=\int_0^1\frac{dx}{(x+\frac{1}{2})^2+\frac{3}{4}}=\frac{4}{3}\int_0^1\frac{dx}{\left(\frac{2x+1}{\sqrt{3}}\right)^2+1}=\frac{2}{\sqrt{3}}\int_{\frac{1}{\sqrt{3}}}^\sqrt{3}\frac{du}{u^2+1}\\&=\frac{2}{\sqrt{3}}\left[\arctan u\right]_{\frac{1}{\sqrt{3}}}^\sqrt{3}=\frac{\pi\sqrt{3}}{9}\end{align}

We can calculate the other integral by the same method.

share|improve this answer
    
Thank you. This is exactly what I need. Niceeee! –  Peter Wormy May 5 '13 at 21:06
    
You're welcome. –  Sami Ben Romdhane May 5 '13 at 21:06
    
I saw another answer but it disappeared. That answer was correct but wrongly explained. There one must apply l'Hopital, not to factorize S(n) because S(n) can be viewed as a simple variable that tends to -beta. –  Peter Wormy May 5 '13 at 21:09
    
could you add the variable change you used? –  Peter Wormy May 5 '13 at 21:22
    
@PeterWormy Ok I'll edit my answer. –  Sami Ben Romdhane May 5 '13 at 21:25

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.