Find sum of series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{2n - 1}$

Question

Given series $$\sum_{n=1}^{\infty}\frac{\cos(nx)}{2n-1}$$

how can we find the sum of given series?

Update:

This task is from calculus workbook we got in class, so I can't give a valid source where the task came from. I've tried to change $$\cos$$ to $$z^n$$ so I would have a power series and it would be possible to create a geometric progression using it, and then I would find the real part of it. Unfortunately, I can't implement this idea because after that I don't know how to get rid of complex numbers

Answer 1

For $|z|\leq 1$ and $z\neq -1$ , you have

$$\log(1+z)=\sum_{r=1}^{\infty}\frac{(-1)^{r-1}z^{r}}{r}$$

So $\frac{1}{2}\bigg(\log(1+z)-\log(1-z)\bigg)=\sum_{r=1}^{\infty}\frac{z^{2r-1}}{2r-1}$

And hence $\frac{z}{2}\bigg(\log(1+z)-\log(1-z)\bigg)=\sum_{r=1}^{\infty}\frac{z^{2r}}{2r-1}$

Now let $z=e^{\frac{ix}{2}}$ to get,

$$\Re\bigg(\frac{e^{\frac{ix}{2}}}{2}\bigg(\log(1+e^{\frac{ix}{2}})-\log(1-e^{\frac{ix}{2}})\bigg)\bigg)=\sum_{k=1}^{\infty}\frac{\cos(kx)}{2k-1}$$

Now use the principal branch of complex logarithm to get the real part.

i.e. define $\log(re^{i\theta})=\ln(r)+i\theta\,\,,\theta \in (-\pi,\pi]$

The reason that we can chose the principal branch of complex logarithm arbitrarily is due to the trivial reason that the sum itself converges and it is a real sum . So whatever branch of logarithm you chose, you would see that the value is consistent. So I'll leave you to evaluate the moduluses and computing the real part. It could be a bit tedious though.

Answer 2

Without complex numbers, one can give a solution based on combining a few well-known Fourier series (which are easily confirmed by integration by parts as shown Here):

$$\sum_{n=1}^{\infty}\frac{\cos((2n-1)x)}{2n-1}=-\frac{\log \tan (|x|/2)}{2}, |x| < \pi, x \ne 0$$ $$\sum_{n=1}^{\infty}\frac{\sin((2n-1)x)}{2n-1}=\frac{\pi}{4} \text {sign }x, |x| < \pi, x \ne 0$$

Letting for now $0<x<\pi$ and multiplying the first equation by $\cos x$ and the second by $\sin x$ and subtracting, one gets:

$$\sum_{n=1}^{\infty}\frac{\cos(2nx)}{2n-1}=-\cos x\frac{\log \tan (x/2)}{2}-\frac{\pi \sin x}{4}, 0< x < \pi$$

Letting now $x \to x/2$ so indeed $0< x < 2\pi$ we have:

$$\sum_{n=1}^{\infty}\frac{\cos(nx)}{2n-1}=-\frac{1}{2}\cos\left(\frac{x}{2}\right)\log\tan\left(\frac{x}{4}\right)-\frac{\pi}{4}\sin\left(\frac{x}{2}\right), 0< x < 2\pi$$

Answer 3

Using the substitution, $\displaystyle{\cos(nx)=\frac{e^{inx}+e^{-inx}}{2}}$ $$\begin{align*} \sum_{n=1}^\infty\frac{\cos(nx)}{2n-1}&=\frac{1}{2}\sum_{n=1}^\infty\frac{e^{inx}+e^{-inx}}{2n-1} \\ &=\frac{1}{2}\sum_{n=0}^{\infty}\frac{e^{ix(n+1)}+e^{-ix(n+1)}}{2n+1}\\ &=\frac{1}{2}\left(\sum_{n=0}^\infty\frac{e^{inx+ix}}{2n+1}+\sum_{n=0}^\infty\frac{e^{-inx-ix}}{2n+1}\right)\\ &=\frac{1}{2}\left(e^{ix/2}\sum_{n=0}^\infty\frac{e^{inx+ix/2}}{2n+1}+e^{-ix/2}\sum_{n=0}^{\infty}\frac{e^{-inx-ix/2}}{2n+1}\right)\\ &=\frac{1}{2}\left(e^{ix/2}\sum_{n=0}^\infty\frac{e^{(2n+1)ix/2}}{2n+1}+e^{-ix/2}\sum_{n=0}^\infty\frac{e^{-(2n+1)ix/2}}{2n+1}\right) \end{align*}$$ Now by the Maclaurin series for the inverse hyperbolic tangent we find the closed form solution, $$\tanh^{-1}(x)=\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}$$$$\begin{align*} \sum_{n=1}^\infty\frac{\cos(nx)}{2n-1}&=\frac{1}{2}\left(e^{ix/2}\sum_{n=0}^\infty\frac{e^{(2n+1)ix/2}}{2n+1}+e^{-ix/2}\sum_{n=0}^\infty\frac{e^{-(2n+1)ix/2}}{2n+1}\right)\\ &=\boxed{\frac{1}{2}\left(e^{ix/2}\tanh^{-1}(e^{ix/2})+e^{-ix/2}\tanh^{-1}(e^{-ix/2})\right)} \end{align*}$$