Tangent series exercise

Evaluate the following value of tangent function, if it exists:

$$\tan \left[ \sum_{r = 1}^{\infty} \arctan \left( \displaystyle \frac{4}{4r^2 + 3} \right) \right]$$

First of all, the expression must be decomposed into several parts. The series is considered first. There can be no clues about its convergence. However, a fundamental relation can be recalled here: the $\arctan$ addition formula. It will be taken with a negative value for the second angle, so it becomes:

$$\arctan(\alpha) - \arctan (\beta) = \arctan \left( \displaystyle \frac{\alpha - \beta}{1 + \alpha \beta} \right)$$

Of course, it must be $\alpha \beta \neq -1$ (otherwise, the denominator will not exist), and this result is given $\mod \pi$. In fact, being $\pi$ the period of the tangent function, the angle $\arctan(x)$ generates the same tangent value of $\arctan(x) + m\pi, m \in \mathbb{Z}$, so they are indistinguishable.

Even if this relation does not immediately seem related to this problem, it could be the only chance to rewrite the above expression in a more convenient way. So, a rearrangement attempt must be performed, to verify if the $\arctan$ addition formula can actually be applied.

Instead of the identification of $\alpha$ and $\beta$, it could be more convenient to directly connect the expression $4 / (4r^2 + 3)$ to $(\alpha - \beta) / (1 + \alpha \beta)$. Several attempts are necessary. First,

$$\displaystyle \frac{4}{4r^2 + 3} = \frac{1}{r^2 + \displaystyle \frac{3}{4}}$$

The denominator has currently no unity value (as it should instead), however the fraction $3 / 4$ can always be rewritten as $1 - (1/2)^2$:

$$\displaystyle \frac{4}{4r^2 + 3} = \frac{1}{1 + r^2 - \displaystyle \left( \frac{1}{2} \right)^2}$$

This produces two benefits:

• first, a unity value appears in the denominator;
• second, the chance to exploit the product of sum and difference of two numbers appears.

This is exactly what was needed, because the product $\alpha \beta$ must be contained in the denominator:

$$r^2 - \displaystyle \left( \frac{1}{2} \right)^2 = \left( r + \frac{1}{2} \right)\left( r - \frac{1}{2} \right)$$

Remembering the $\arctan$ addition formula, $\displaystyle \left( r + \frac{1}{2} \right)$ is candidate to be $\alpha$ and $\displaystyle \left( r - \frac{1}{2} \right)$ is candidate to be $\beta$. If this is true, also the numerator should be rearranged such that it contains the difference $\alpha - \beta$. It currently does not. However:

$$1 = \displaystyle \left( r + \frac{1}{2} \right) - \left( r - \frac{1}{2} \right)$$

So,

$$\displaystyle \frac{4}{4r^2 + 3} = \displaystyle \frac{\left( r + \displaystyle \frac{1}{2} \right) - \displaystyle \left( r - \displaystyle \frac{1}{2} \right)}{1 + \left( r + \displaystyle \frac{1}{2} \right)\left( r - \displaystyle \frac{1}{2} \right)}$$

This is exactly the form $(\alpha - \beta) / (1 + \alpha \beta)$. Then, the $\arctan$ addition formula can be applied to the series:

$$\sum_{r = 1}^{\infty} \arctan \left( \displaystyle \frac{4}{4r^2 + 3} \right) = \sum_{r = 1}^{\infty} \arctan \left[ \displaystyle \frac{\left( r + \displaystyle \frac{1}{2} \right) - \displaystyle \left( r - \displaystyle \frac{1}{2} \right)}{1 + \left( r + \displaystyle \frac{1}{2} \right)\left( r - \displaystyle \frac{1}{2} \right)} \right] =\\ = \sum_{r = 1}^{\infty} \left[ \arctan \displaystyle \left( r + \frac{1}{2} \right) - \arctan \displaystyle \left( r - \frac{1}{2} \right) \right]$$

The series can be still unknown, as well as its convergence. Each of its members is composed by an actual difference of values. One well-known kind of series which is composed the same way is the telescoping series. It has the following structure:

$$\sum_{n = 1}^{N} (a_{n + 1} - a_n) = a_{N + 1} - a_1$$

Each term is composed by two addends. When the series is expanded, only the first and the last addends get not erased in the algebraic sum. If they are finite values, the series converges.

The above series can be rewritten as follows:

$$\sum_{r = 1}^{\infty} \left[ \arctan \displaystyle \left( r + \frac{1}{2} \right) - \arctan \displaystyle \left( r - \frac{1}{2} \right) \right] =\\ = \sum_{r = 1}^{\infty} \left\{ \arctan \displaystyle \left[ (r + 1) - \frac{1}{2} \right] - \arctan \displaystyle \left( r - \frac{1}{2} \right) \right\}$$

It is a telescoping series. In this case, the last addend has not a finite index $N$: it is

$$\lim_{r \to \infty} \arctan \displaystyle \left[ (r + 1) - \frac{1}{2} \right] = \displaystyle \frac{\pi}{2}$$

Note that this is not an ordinary angle, but a limit angle value. It is finite. So,

$$\sum_{r = 1}^{\infty} \left\{ \arctan \displaystyle \left[ (r + 1) - \frac{1}{2} \right] - \arctan \displaystyle \left( r - \frac{1}{2} \right) \right\} = \displaystyle \frac{\pi}{2} - \arctan \left( \displaystyle \frac{1}{2} \right)$$

The series converges; $\arctan(1/2)$ is not a well-known angle value (easily writeable angles which generate the value $1/2$ for sine and cosine are available, but not for tangent).

The initial expression becomes:

$$\tan \left[ \displaystyle \frac{\pi}{2} - \arctan \left( \displaystyle \frac{1}{2} \right) \right]$$

First, this is a value different from (and smaller than) $\pi / 2$, so it is acceptable and included in the domain of tangent.

According to the trigonometric identities,

$$\tan \left( \displaystyle \frac{\pi}{2} - \alpha \right) = \cot \alpha$$ $$\tan \left[ \displaystyle \frac{\pi}{2} - \arctan \left( \displaystyle \frac{1}{2} \right) \right] = \cot \left[ \arctan \left( \displaystyle \frac{1}{2} \right) \right] = \displaystyle \frac{1}{\tan \left[ \arctan \left( \displaystyle \frac{1}{2} \right) \right]} = \frac{1}{\displaystyle \frac{1}{2}} = 2$$

which is still a limit value (due to $r \to \infty$) and represents the final result.

---

The difference identity for the tangent function could also alternatively be recalled here. If directly using the angle $\pi / 2$, this identity would explicitly include $\tan \left( \displaystyle \frac{\pi}{2} \right)$, which is a meaningless expression. However, being the angle $\pi / 2$ only a limit value, we are allowed to use the limit of this identity. The identity is indeed applicable to all the angles, except $\pi / 2$ (and its integer multiples).

$$\lim_{x \to \pi / 2} \tan \left[ x - \arctan \left( \displaystyle \frac{1}{2} \right) \right] = \frac{\tan x - \tan \left[ \arctan \left( \displaystyle \frac{1}{2} \right) \right]}{1 + \tan x \tan \left[ \arctan \left( \displaystyle \frac{1}{2} \right) \right]}$$

Only the terms going to infinite are not negligible: $\tan x$ in the numerator and $\tan x \tan \left[ \arctan \left( \displaystyle \frac{1}{2} \right) \right]$ in the denominator. The result is

$$\lim_{x \to \pi / 2} \tan \left[ x - \arctan \left( \displaystyle \frac{1}{2} \right) \right] = \frac{1}{\tan \left[ \arctan \left( \displaystyle \frac{1}{2} \right) \right]} = 2$$

–

This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License