Question

Vă rog frumos dacă mă puteți ajuta!

Answer 1

Aplicam L'Hopital deoarece avem cazul 0/0.

$\lim\limits_{x\to 0}\dfrac{1}{\sin^2 (nx)}\cdot \sum\limits_{k=1}^{2n}(-1)^{k+1}\cos(kx)=\\ \\ = \lim\limits_{x\to 0} \dfrac{\cos 0 - \cos 0+\cos 0-\cos 0+...+\cos 0 - \cos 0}{0}\overset{\frac{0}{0}}{=}$

$\overset{\frac{0}{0}}{=}\lim\limits_{x\to 0}\dfrac{\sum\limits_{k=1}^{2n}(-1)^{k+1}\cdot [-\sin(kx)]\cdot k }{2\sin(nx)\cdot\cos (nx)\cdot n}= \lim\limits_{x\to 0}\dfrac{-\sum\limits_{k=1}^{2n} (-1)^{k+1}k\sin (kx)}{2n\sin(nx)\cos(nx)}\overset{\frac{0}{0}}{=}\\ \\ \overset{\frac{0}{0}}{=}\lim\limits_{x\to 0}\dfrac{\sum\limits_{k=1}^{2n}(-1)^{k}k\cos(kx)\cdot k}{2n\cos^2(nx)\cdot n-2n\sin^2 (nx)\cdot n}=\\ \\ \\=\dfrac{-1^2+2^2-3^2+...-(2n-1)^2+(2n)^2}{2n^2}=$

$= \dfrac{(2-1)(2+1)+(4-3)(4+3)+..+\Big(2n-(2n-1)\Big)(2n+2n-1)}{2n^2} = \\ \\ = \dfrac{1\cdot 3+1\cdot 7+1\cdot 11+...+1\cdot(4n-1) }{2n^2} = \\ \\ =\dfrac{(1+2)+(3+4)+(5+6)+...+(2n-1+2n)}{2n^2} = \\ \\ = \dfrac{2n(2n+1)}{2n^2}= \boxed{\dfrac{2n+1}{n}}$