# Infinite Sum of Series

So I was given this question $$T_n = \sum _ {k=0}^{ n-1} \frac{n}{n^2+kn+ k^2}$$ And $$S_n = \sum _{ k=1}^n \frac{n}{n^2+kn+ k^2}$$ We were asked wether $$T_n$$ or$$S_n$$is$$\gt$$or$$\lt \frac{π}{3\sqrt3}$$

So what I have deduced is that when $$lim _ { n \to \infty }$$this is sum of infinite series and upon calculating we get $$\int_{x=0}^1 \frac{1}{x^2 +x+1} dx$$ Which is $$\frac{π}{3√3}$$ but this happens with both the integrals , how to decide the sign ??

• What didn't you understand? The change of summation into integral? – Tojrah May 19 at 4:08
• Try changing the limits of the summation $T_n$ by substituting $n-1=t$ or something similar, maybe? – ExtremeRaider May 19 at 4:14

Hint: Since $$\dfrac{1}{x^2+x+1}$$ is decreasing over $$x \in [0,1]$$, we have $$\int_{\tfrac{k-1}{n}}^{\tfrac{k}{n}}\dfrac{1}{\left(\tfrac{k}{n}\right)^2+\left(\tfrac{k}{n}\right)+1}\,dx \le \int_{\tfrac{k-1}{n}}^{\tfrac{k}{n}}\dfrac{1}{x^2+x+1}\,dx \le \int_{\tfrac{k-1}{n}}^{\tfrac{k}{n}}\dfrac{1}{\left(\tfrac{k-1}{n}\right)^2+\left(\tfrac{k-1}{n}\right)+1}\,dx$$ i.e. $$\dfrac{\tfrac{1}{n}}{\left(\tfrac{k}{n}\right)^2+\left(\tfrac{k}{n}\right)+1}\le \int_{\tfrac{k-1}{n}}^{\tfrac{k}{n}}\dfrac{1}{x^2+x+1}\,dx \le \dfrac{\tfrac{1}{n}}{\left(\tfrac{k-1}{n}\right)^2+\left(\tfrac{k-1}{n}\right)+1}.$$
Now, what do you get when you sum each part of the inequality from $$k = 1$$ to $$n$$?
If it is true that they both tend to the quantity $$\frac {π}{3\sqrt 3}=L$$ as $$n\to\infty,$$ then since they both have positive terms, it follows that they are both less than $$L$$ when you truncate them at $$n.$$
• All you have proved is $S_n \gt T_n$ I want their relation with π /3√3 – user232243 May 19 at 4:38