Question Number 28614 by abdo imad last updated on 27/Jan/18
$${find}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{sin}\left({nx}\right)}{{n}}. \\ $$$$ \\ $$
Commented by abdo imad last updated on 30/Jan/18
![let developp the function f(x)=(x/2)periodic by 2π and odd f(x) = Σ_(n=1) ^∞ a_n sin(nx) and a_n = (2/(2π)) ∫_([T]) (x/2) sin(nx)dx =(1/π) ∫_0 ^π x sin(nx)dx and by parts π a_n =((−x)/n)cos(nx)]_0 ^π − ∫_0 ^π ((−1)/n)cos(nx)dx =((−π)/n)(−1)^n +(1/n^2 )[sin(nx)]_0 ^π = ((π(−1)^(n−1) )/n) so a_n =(((−1)^(n−1) )/n) and (x/2)= Σ_(n=1) ^∞ (((−1)^(n−1) )/n) sin(nx) let do thech.=π−t so ((π−t)/2)=Σ_(n=1) ^∞ (((−1)^(n−1) )/n)sin(nπ−nt) but sin(nπ−nt)=sin(nπ)cos(nt)−cos(nπ)sin(nt) =(−1)^(n−1) sin(nt)⇒ ((π−t)/2) =Σ_(n=1) ^∞ ((sin(nt))/n) finally we have Σ_(n=1) ^∞ ((sin(nx))/n) = ((π−x)/2) .](https://www.tinkutara.com/question/Q28761.png)
$${let}\:{developp}\:{the}\:{function}\:{f}\left({x}\right)=\frac{{x}}{\mathrm{2}}{periodic}\:{by}\:\mathrm{2}\pi\:{and}\:{odd} \\ $$$${f}\left({x}\right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} {a}_{{n}} {sin}\left({nx}\right)\:{and}\:{a}_{{n}} =\:\frac{\mathrm{2}}{\mathrm{2}\pi}\:\int_{\left[{T}\right]} \:\frac{{x}}{\mathrm{2}}\:{sin}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{x}\:{sin}\left({nx}\right){dx}\:{and}\:{by}\:{parts} \\ $$$$\left.\pi\:{a}_{{n}} =\frac{−{x}}{{n}}{cos}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:−\:\int_{\mathrm{0}} ^{\pi} \frac{−\mathrm{1}}{{n}}{cos}\left({nx}\right){dx} \\ $$$$=\frac{−\pi}{{n}}\left(−\mathrm{1}\right)^{{n}} \:\:+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left[{sin}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:=\:\frac{\pi\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\:{so} \\ $$$${a}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\:{and}\: \\ $$$$\frac{{x}}{\mathrm{2}}=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{sin}\left({nx}\right)\:\:{let}\:{do}\:{thech}.=\pi−{t}\:{so} \\ $$$$\frac{\pi−{t}}{\mathrm{2}}=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{sin}\left({n}\pi−{nt}\right)\:\:{but} \\ $$$${sin}\left({n}\pi−{nt}\right)={sin}\left({n}\pi\right){cos}\left({nt}\right)−{cos}\left({n}\pi\right){sin}\left({nt}\right) \\ $$$$=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {sin}\left({nt}\right)\Rightarrow\:\frac{\pi−{t}}{\mathrm{2}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{sin}\left({nt}\right)}{{n}}\:{finally}\:{we}\:{have} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{sin}\left({nx}\right)}{{n}}\:=\:\frac{\pi−{x}}{\mathrm{2}}\:\:. \\ $$$$ \\ $$
Commented by abdo imad last updated on 30/Jan/18
$${the}\:{convergence}\:{of}\:{this}\:{serie}\:{is}\:{assured}\:{by}\:{Abel}\:{Dirichlet} \\ $$$${theorem}\:{let}\:{remember}\:{this}\:{theorem}.{if}\:\Sigma\:{a}_{{n}} \left({x}\right){v}_{{n}} \left({x}\right){is}\:{a}\: \\ $$$${serie}\:{of}\:{function}\:\:\left(\:\:{a}_{{n}} >\mathrm{0}\:{and}\:{v}_{{n}} >\mathrm{0}\right)/\:{a}_{{n}\:} {decrease}\:{and}\:{a}_{{n}} \left({x}\right)_{{n}\rightarrow\infty} \rightarrow\mathrm{0} \\ $$$${and}\exists\:{m}>\mathrm{0}\:/\:\mid\sum_{{k}={n}_{\mathrm{0}} } ^{{n}} \:{v}_{{k}} \left({x}\right)\mid\leqslant{m}\:{so}\:{the}\:{serie}\:{converges}. \\ $$