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Question Number 33131 by prof Abdo imad last updated on 11/Apr/18

$$1)find\sum _{n=1}^{\mathrm{\infty }}\frac{e^{inx}}{n(n+1)}$$$$2)findthevalueof\sum _{n⩾1}\frac{sin\left(nx\right)}{n(n+1)}$$$$and\sum _{n⩾1}\frac{cos\left(nx\right)}{n(n+1)}.$$

Commented by prof Abdo imad last updated on 12/Apr/18

$$letfindS\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n(n+1)}theradiusof$$$$convergrnceis1andforx=\stackrel{-}{+}1theserieis$$$$alsoconvergentwehavefor\mid x\mid <1$$$$S\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{n}-\frac{1}{n+1})x^n=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n+1}$$$$but\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}=-ln(1-x)$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n+1}=\sum _{n=2}^{\mathrm{\infty }}\frac{x^{n-1}}{n}=\frac{1}{x}\sum _{n=2}^{\mathrm{\infty }}\frac{x^n}{n}$$$$=\frac{1}{x}(-ln(1-x)-x)=-\frac{1}{x}ln(1-x)-1\Rightarrow$$$$S\left(x\right)=-ln(1-x)+\frac{1}{x}ln(1-x)+1$$$$=(-1+\frac{1}{x})ln(1-x)+1\Rightarrow$$$$S\left(x\right)=\frac{1-x}{x}ln(1-x)+1letchangexper{e}^{ix}$$$$weget\sum _{n=1}^{\mathrm{\infty }}\frac{e^{inx}}{n(n+1)}=\frac{1-e^{ix}}{e^{ix}}ln(1-e^{ix})+1$$$$=(e^{-ix}-1)ln(1-e^{ix})+1$$$$2)\sum _{n=1}^{\mathrm{\infty }}\frac{sin\left(nx\right)}{n(n+1)}=Im\left(S\left(e^{ix}\right)\right)letfindit$$$$(e^{-ix}-1)=cosx-isinx-1$$$$$$$$=-2{sin}^2\left(\frac{x}{2}\right)-2isin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\right)$$$$=-2isin\left(\frac{x}{2}\right)(cos\left(\frac{x}{2}\right)-isin\left(\frac{x}{2}\right))$$$$=-2i{e}^{-i\frac{x}{2}}$$$$ln(1-e^{ix})=ln(1-cosx-isinx)$$$$=ln(2{sin}^2\left(\frac{x}{2}\right)-2isin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\right))$$$$=ln(-2isin\left(\frac{x}{2}\right)(cos\left(\frac{x}{2}\right)+isin\left(\frac{x}{2}\right))$$$$=ln(-i)+ln(2sin\left(\frac{x}{2}\right)+ln\left(e^{i\frac{x}{2}}\right)$$$$=-i\frac{\pi }{2}+ln\left(2sin\left(\frac{x}{2}\right)\right)+\frac{ix}{2}$$$$=ln\left(2sin\left(\frac{x}{2}\right)\right)+i\frac{x-\pi }{2}$$$$Im\left(S\left(e^{ix}\right)\right)=-2i{e}^{-i\frac{x}{2}}(ln(2sin\left(\frac{x}{2}\right)+i\frac{x-\pi }{2})$$$$=-2i(cos\left(\frac{x}{2}\right)-isin\left(\frac{x}{2}\right))(ln(2sin\left(\frac{x}{2}\right)+i\frac{x-\pi }{2})$$$$=-2i)(cos\left(\frac{x}{2}\right)ln(2sin\left(\frac{x}{2}\right)+i\frac{x-\pi }{2}cos\left(\frac{x}{2}\right)$$$$-isin\left(\frac{x}{2}\right)ln(2sin\left(\frac{x}{2}\right)+\frac{x-\pi }{2}sin\left(\frac{x}{2}\right))$$$$=-2icos\left(\frac{x}{2}\right)ln(2sin\left(\frac{x}{2}\right)+(x-\pi )cos\left(\frac{x}{2}\right)$$$$-2sin\left(\frac{x}{2}\right)ln(2sin\left(\frac{x}{2}\right)+i(\pi -x)sin\left(\frac{x}{2}\right)\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{sin\left(nx\right)}{n(n+1)}=(\pi -x)sin\left(\frac{x}{2}\right)-sin\left(\frac{x}{2}\right)ln\left(2sin\left(\frac{x}{2}\right)\right)$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{cos\left(nx\right)}{n(n+1)}=(x-\pi )cos\left(\frac{x}{2}\right)-2sin\left(\frac{x}{2}\right)ln\left(2sin\left(\frac{x}{2}\right)\right)+1.$$$$$$

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