Finding the Sum of the Given Cosine Series

We are asked to find the sum of series given by:

\(\cos \frac{\pi}{3}+\frac{1}{2} \cos \frac{2 \pi}{3} + \frac{1}{3} \cos \frac{3 \pi}{3} + \ldots\) up to infinity.

This is an infinite series that can be written in the summation notation as:

\(\sum_{n=1}^{\infty} \frac{1}{n} \cos \left(\frac{n \pi}{3}\right)\)

This is a specific instance of a trigonometric or cosine series.

Connecting the Cosine Series to Complex Logarithmic Series

This type of series often relates to the real or imaginary parts of complex series expansions, particularly the logarithmic series. We utilize the properties of complex numbers for this approach. Recall the Taylor series expansion for \(\log(1-z)\) for \(|z| \le 1, z \ne 1\):

\(\log(1-z) = -\sum_{n=1}^{\infty} \frac{z^n}{n}\)

Let's consider a complex number \(z\) on the unit circle, \(z = e^{i \theta}\). Substituting this into the logarithmic series:

\(\log(1 - e^{i \theta}) = -\sum_{n=1}^{\infty} \frac{(e^{i \theta})^n}{n} = -\sum_{n=1}^{\infty} \frac{e^{i n \theta}}{n}\)

Using Euler's formula, \(e^{i n \theta} = \cos(n \theta) + i \sin(n \theta)\), we get:

\(\log(1 - e^{i \theta}) = -\sum_{n=1}^{\infty} \frac{\cos(n \theta) + i \sin(n \theta)}{n}\)

\(\log(1 - e^{i \theta}) = -\sum_{n=1}^{\infty} \frac{\cos(n \theta)}{n} - i \sum_{n=1}^{\infty} \frac{\sin(n \theta)}{n}\)

The real part of this equation is \(\text{Re}[\log(1 - e^{i \theta})] = -\sum_{n=1}^{\infty} \frac{\cos(n \theta)}{n}\). Thus, the sum of series \(\sum_{n=1}^{\infty} \frac{\cos(n \theta)}{n}\) is equal to \(-\text{Re}[\log(1 - e^{i \theta})]\).

Calculating the Sum of the Specific Cosine Series

Comparing our given cosine series \(\sum_{n=1}^{\infty} \frac{1}{n} \cos \left(\frac{n \pi}{3}\right)\) with the general form \(\sum_{n=1}^{\infty} \frac{\cos(n \theta)}{n}\), we identify \(\theta = \frac{\pi}{3}\). We need to calculate \(-\text{Re}[\log(1 - e^{i \pi/3})]\) to find the sum of series. This involves calculations with complex numbers.

First, let's evaluate \(1 - e^{i \pi/3}\). Using Euler's formula:

\(e^{i \pi/3} = \cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right)\)

We know the values for \(\cos(\pi/3)\) and \(\sin(\pi/3)\):

So, \(e^{i \pi/3} = \frac{1}{2} + i \frac{\sqrt{3}}{2}\).

Now, calculate \(1 - e^{i \pi/3}\):

\(1 - e^{i \pi/3} = 1 - \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = \frac{1}{2} - i \frac{\sqrt{3}}{2}\)

Evaluating the Complex Logarithm for the Sum of Series

We need to find \(\log\left(\frac{1}{2} - i \frac{\sqrt{3}}{2}\right)\). Let \(w = \frac{1}{2} - i \frac{\sqrt{3}}{2}\). To evaluate the logarithm of a complex number \(w = x + iy\), we convert it to polar form, \(w = r e^{i \phi}\), where \(r = |w|\) is the magnitude and \(\phi = \arg(w)\) is the argument (angle).

The magnitude \(r\) is:

\(r = \left|\frac{1}{2} - i \frac{\sqrt{3}}{2}\right| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1\)

The argument \(\phi\) is the angle whose tangent is \(\frac{-\sqrt{3}/2}{1/2} = -\sqrt{3}\). Since the real part (\(1/2\)) is positive and the imaginary part (\(-\sqrt{3}/2\)) is negative, the complex number lies in the fourth quadrant. The principal argument is \(\phi = -\frac{\pi}{3}\).

So, \(w = 1 \cdot e^{-i \pi/3}\).

The principal value of the complex logarithm \(\log(w)\) is given by \(\log(r) + i \phi\):

\(\log\left(\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) = \log(1) + i \left(-\frac{\pi}{3}\right) = 0 - i \frac{\pi}{3} = -i \frac{\pi}{3}\)

This result comes from the logarithmic series evaluation.

Final Sum of the Infinite Series

The sum of series is \(-\text{Re}[\log(1 - e^{i \pi/3})]\). We found that \(\log(1 - e^{i \pi/3}) = -i \frac{\pi}{3}\). The real part of \(-i \frac{\pi}{3}\) is 0.

\(\text{Re}\left[-i \frac{\pi}{3}\right] = \text{Re}\left[0 - i \frac{\pi}{3}\right] = 0\)

Therefore, the sum of series for the given infinite series is:

\(-\text{Re}\left[-i \frac{\pi}{3}\right] = -(0) = 0\)

The value of the given series is 0.