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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 ??

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

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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 $n.$

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  • $\begingroup$ All you have proved is $S_n \gt T_n$ I want their relation with π /3√3 $\endgroup$ – user232243 May 19 at 4:38
  • $\begingroup$ @user232243 Please see edit. $\endgroup$ – Allawonder May 19 at 4:47

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