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

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 $$(1+\cos x)\ln\left(2\cos\frac{x}{2}\right)+\frac{1}{2}x\sin x - 1$$.

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