Riemann sum of Integral

Real Analysis
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jacks
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Riemann sum of Integral

#1

Post by jacks »

Let \(\displaystyle S_{n} = \sum_{k=1}^{n}\frac{n}{n^2+nk+k^2}\) and \(\displaystyle T_{n} = \sum_{k=0}^{n-1}\frac{n}{n^2+nk+k^2}\) for \(n=1,2,3,4.....\) Then which one is Right

\(\displaystyle (a):: S_{n}<\frac{\pi}{3\sqrt{3}}\)

\(\displaystyle (b):: S_{n}>\frac{\pi}{3\sqrt{3}}\)

\(\displaystyle (c):: T_{n}<\frac{\pi}{3\sqrt{3}}\)

\(\displaystyle (d):: T_{n}>\frac{\pi}{3\sqrt{3}}\)
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Grigorios Kostakos
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Re: Riemann sum of Integral

#2

Post by Grigorios Kostakos »

The function \(f(x)=\frac{1}{1+x+x^2}\) is Riemann integrable and strictly decreasing in the interval \([0,1]\). If \({\cal{D}}_{n}=\bigl\{{\bigl[{\frac{k-1}{n},\frac{k}{n}}\bigr]\;|\;k=1,2\ldots,n}\bigr\}\,,\; n\in{\mathbb{N}}\,, \) is a partition of the interval \([0,1]\), then \[\displaystyle S_{n} = \sum_{k=1}^{n}\frac{n}{n^2+nk+k^2}= \frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}+\bigl({\frac{k}{n}}\bigr)^2}\,,\quad n\in{\mathbb{N}}\,,\]
and
\[\displaystyle T_{n} = \sum_{k=0}^{n-1}\frac{n}{n^2+nk+k^2}= \frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{1+\frac{k}{n}+\bigl({\frac{k}{n}}\bigr)^2}\,,\quad n\in{\mathbb{N}}\,,\] are the Lower Sum and the Upper Sum, respectively, of the function \(f\) in the interval \([0,1]\). So we have that
\[S_{n}< M_{n}< T_{n}\quad n\in{\mathbb{N}}\,,\]
where \(M_{n}\) is the Middle Sum of the function \(f\) in the interval \([0,1]\). Because \[\displaystyle\mathop{\lim}\limits_{n\to+\infty}{M_{n}}=\int_{0}^{1}{\frac{1}{1+x+x^2}\,dx}=\frac{\pi\sqrt{3}}{9}\,,\]
we have that \[S_{n}< \frac{\pi\sqrt{3}}{9}< T_{n}\] for every \(n\in{\mathbb{N}}\).
Grigorios Kostakos
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