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

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

• Are you sure that you are required to "evaluate" the sum and not check if the sum converges? Commented Apr 26, 2023 at 11:01
• Follow the same ideas as here. Commented Apr 26, 2023 at 11:05
• It is divergent when $x=0$. Commented Apr 26, 2023 at 11:20
• If this is just a calculus workbook, nothing advanced, then I would expect $x=\pi$ in that series. Commented Apr 26, 2023 at 11:25

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.

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 $$0 and multiplying the first equation by $$\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:

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

• (+1) Hi Conrad! How are you? It's good to see you're still around. I hope that you don't mind the edit with an added reference, but thought it might be useful. Commented Apr 26, 2023 at 15:36
• no problem - the above are standard Fourier series and can be found in many places, while once known (that is the hard part imho and that's why having some tables of such is handy) they are easily checked; the Tolstov book on the subject has an extensive list in chapter 5 but the standard Tables books (eg Gradshteyn/Rhyzik various editions) also have them for example; see store.doverpublications.com/0486633179.html Commented Apr 26, 2023 at 15:45

Using the substitution, $$\cos(nx)=({e^{inx}+e^{-inx}})/{2}$$ \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*}

• Is there an expression not involving complex number at the end ? It would be more naturel given that the sum is real (for real $x$ at least). Commented Apr 26, 2023 at 11:29
• @Lelouch. Looking at math.stackexchange.com/questions/2541236/…, I am confident there is.
– bob
Commented Apr 26, 2023 at 11:37
• The expression you gave only applies when $|x|< 1$, while the values you applied $\tanh^{-1}$ to are all of modulus $1$, so you'd need to justify the fact that you can still write what you wrote? Haven't used hyperbolic functions that frequently though, so maybe that formula still works for some or almost all $|x| = 1$, I wouldn't know. One problem though for example is that every series here diverges for $x = 0$ for example... Commented Apr 26, 2023 at 11:41
• Seeing $i$ in a real question is irritating. Commented Apr 26, 2023 at 11:52
• @bob I followed your link (Jack D'Aurizio) and found $-\frac{\pi}{4}\sin\left(\frac{x}{2}\right)+\frac{1}{2}\cos\left(\frac{x}{2}\right)\log\cot\left(\frac{x}{4}\right)$. Correct? Commented Apr 26, 2023 at 13:16