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Find the sum of the series: $$ \sum_{n=1}^\infty (-1)^{n-1}\frac{\cos nx}{n(n+1)} $$

My attempt: $$ \begin{aligned} &\sum_{n=1}^\infty (-1)^{n-1}\frac{\cos nx +i\sin nx}{n(n+1)}=\\ &=\sum_{n=1}^\infty (-1)^{n-1}\left(\frac{(\cos x +i\sin x)^n}{n}-\frac{(\cos x +i\sin x)^n}{n+1}\right)=\\ &=[\cos x + i\sin x = t]=\sum_{n=1}^\infty (-1)^{n-1}\left(\frac{t^n}{n}-\frac{t^n}{n+1}\right)=\dots=\\ &=\ln(1+t)+\frac{1}{t}\left(-\ln|1+t|+t\right) \end{aligned} $$ But I don't know how to end the solution. How to ged rid of $i$ inside the logarithms?

P.S. The answer my textbook gives me is $ (1+\cos x)\ln\left(2\cos\frac{x}{2}\right)+\frac{1}{2}x\sin x - 1 $.

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We have that $$\cos((n+1)x)+\cos(nx)=2\cos(x/2)\cos((n+1/2)x)$$ Therefore, we split the given series into a telescopic sum and another convergent series, $$S(x):=\sum_{n=1}^\infty (-1)^{n-1}\frac{\cos(nx)}{n(n+1)}= \sum_{n=1}^\infty \left((-1)^{n-1}\frac{\cos(nx)}{n}-(-1)^{n}\frac{\cos(n+1)x}{n+1}\right)\\ +2\cos(x/2)\text{Re}\left(\sum_{n=1}^\infty\frac{(-1)^{n} (e^{ix})^{n+1/2}}{n+1}\right).$$ Finally, after recalling that $\ln(1+z)=\sum_{n=1}^\infty (-1)^{n-1}\frac{z^n}{n}$ for $|z|\leq 1$ and $z\not=-1$, we find $$\begin{align} S(x)&=\cos(x)+2\cos(x/2)\text{Re}\left(e^{-ix/2}(\ln(1+e^{ix})-e^{ix})\right)\\ &=\cos(x)+\cos(x/2)(\cos(x/2)\ln(2+2\cos(x))+x\sin(x/2)) -2\cos^2(x/2)\\ &=(1+\cos(x))\ln\left(2\cos(x/2)\right)+\frac{x\sin(x)}{2} - 1. \end{align}$$ where $$\ln(1+e^{ix})=\ln(|1+e^{ix}|)+i\text{Arg}(1+e^{ix}) =\frac{1}{2}\ln(2+2\cos(x))+\frac{ix}{2}.$$

P.S. Your textbook answer is valid for $x\in (-\pi,\pi)$.

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  • $\begingroup$ Could you explain how you got next to last line: $\cos(x)+\cos(x/2)(\cos(x/2)\ln(2+2\cos(x))+x\sin(x/2)) -2\cos^2(x/2)$ ? $\endgroup$ – Bonrey Feb 6 at 15:50
  • $\begingroup$ Please see my last line. $\endgroup$ – Robert Z Feb 6 at 15:59
  • $\begingroup$ @Bonrey Is there in your textbook any assumption for $x$? $\endgroup$ – Robert Z Feb 6 at 16:06

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