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Question Number 55991 by bshahid010@gmail.com last updated on 07/Mar/19

Commented by maxmathsup by imad last updated on 07/Mar/19

$$let{S}_n\left(x\right)=\frac{1}{n^2}\sum _{k=1}^m\left[kx\right]wehave\left[kx\right]⩽kx<\left[kx\right]+1\Rightarrow kx-1<\left[kx\right]⩽kx\Rightarrow$$$$\frac{1}{n^2}\sum _{k=1}^m(kx-1)<\frac{1}{n^2}\sum _{k=1}^m\left[kx\right)⩽\frac{1}{n^2}\sum _{k=1}^{m}kxbut$$$$\frac{1}{n^2}\sum _{k=1}^m(kx-1)=x\frac{\sum _{k=1}^{m}k}{n^2}-\frac{m}{n^2}=x\frac{m(m+1)}{2n^2}-\frac{m}{n^2}=A_{n}also$$$$\frac{1}{n^2}\sum _{k=1}^{m}kx=x\frac{m(m+1)}{n}=B_m$$$$ifmisfixed{lim}_{n\to +\mathrm{\infty }}{A}_n=0and{lim}_{n\to +\mathrm{\infty }}{B}_n=0\Rightarrow lim{S}_n\left(x\right)=0$$$$ifmisfunctionofnitsaothersubjectletsupposem\sim n\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{A}_n=\frac{x}{2}={lim}_{n\to +\mathrm{\infty }}{B}_n\Rightarrow {lim}_{n\to +\mathrm{\infty }}{S}_n\left(x\right)=\frac{x}{2}.$$$$$$

Commented by maxmathsup by imad last updated on 07/Mar/19

$$ifn>m\Rightarrow \mathrm{\exists }qintegr/n=m+q\Rightarrow m=n-q\Rightarrow A_n=x\frac{(n-q)(n-q+1)}{2n^2}-\frac{n-q}{n^2}$$$$\to \frac{x}{2}(n\to +\mathrm{\infty })also{B}_n=x\frac{(n-q)(n-q+1)}{2n^2}\to \frac{x}{2}\Rightarrow {lim}_{n\to +\mathrm{\infty }}{S}_n\left(x\right)=\frac{x}{2}$$$$ifn⩽m\Rightarrow m=n+q\Rightarrow \Rightarrow A_n=x\frac{(n+q)(n+q+)}{2n^2}-\frac{n+q}{2n^2}\to \frac{x}{2}$$$$B_n=x\frac{(n+q)(n+q+1)}{2n^2}\to \frac{x}{2}\Rightarrow {lim}_{n\to +\mathrm{\infty }}{S}_n\left(x\right)=\frac{x}{2}$$$$soatallcaseswehave{lim}_{n\to +\mathrm{\infty }}{S}_n\left(x\right)=\frac{x}{2}$$$$erroroftypo{B}_n=x\frac{m(m+1)}{2n^2}$$

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