Trigonometry limit's proof: limx→0sin⁡(x)+sin⁡(2x)+⋯+sin⁡(kx)x=k(k+1)2I tried to split up the fraction and multiple-divide every new fraction with its x factor but didn't work out. ex:limx→0sin⁡(2x)x=limx→0sin⁡(2x)⋅22⋅x=2

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Trigonometry limit&#039;s proof: limx→0sin⁡(x)+sin⁡(2x)+⋯+sin⁡(kx)x=k(k+1)2I tried to split up the fraction and multiple-divide every new fraction with its x factor but didn&#039;t work out. ex:limx→0sin⁡(2x)x=limx→0sin⁡(2x)⋅22⋅x=2

gil001q4wq · Accepted Answer

You are so close. Note thatsin⁡(x)+sin⁡(2x)+⋯+sin⁡(kx)x=sin⁡(x)x+sin⁡(2x)x+⋯+sin⁡(kx)x=sin⁡(x)x+2sin⁡(2x)2x+⋯+ksin⁡(kx)kx→1+2+⋯+k=k(k+1)2as x→0I suspect you may not have been able to finish because you didn&#039;t recognize the identity1+2+⋯+k=k(k+1)2This identity has a very cute proof. Set S=1+2+⋯+k. Adding1+2+⋯+k=Sk+(k−1)+⋯+1=Sgives(k+1)+(k+1)+⋯+(k+1)⏟k times=2STherefore k(k+1)=2S and consequently1+2+⋯+k=S=k(k+1)2

Trigonometry limit's proof: \(\displaystyle\lim_{{{x}\to{0}}}{\frac{{{\sin{{\left({x}\right)}}}+{\sin{{\left({2}{x}\right)}}}+\cdots+{\sin{{\left({k}{x}\right)}}}}}{{{x}}}}={\frac{{{k}{\left({k}+{1}\right)}}}{{{2}}}}\) I tried to split

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