Prove by induction that $\sum_{r=1}^{n} \arctan \left( \frac{1}{2r^2} \right) = \arctan \left( \frac{n}{n+1} \right)$

Question

Prove by induction that $$\sum_{r=1}^{n} \arctan \left( \frac{1}{2r^2} \right) = \arctan \left( \frac{n}{n+1} \right)$$ for $n \in \mathbb{Z^+}$

Let $P(n)$ be the statement $$\sum_{r=1}^{n} \arctan \left( \frac{1}{2r^2} \right) = \arctan \left( \frac{n}{n+1} \right)$$ for $n \in \mathbb{Z^+}$.

Base case, $P(1)$: $$\arctan \left( \frac{1}{2} \right) = \arctan \left( \frac{1}{2} \right)$$

Therefore $P(1)$ is true. Assume that $P(k)$ is true. So,

$$\arctan \left( \frac{1}{2(1)^2} \right) + \cdots + \arctan \left( \frac{1}{2(k)^2} \right) = \arctan \left( \frac{k}{k+1} \right)$$

Prove $P(k+1)$ is true:

\begin{align} &\arctan \left( \frac{1}{2(1)^2} \right) + \cdots + \arctan \left( \frac{1}{2(k)^2} \right) + \arctan \left( \frac{1}{2(k+1)^2} \right) \\ = \ &\arctan \left( \frac{k}{k+1} \right) + \arctan \left( \frac{1}{2(k+1)^2} \right) \end{align}

I'm stuck on what to do with $$\arctan \left( \frac{k}{k+1} \right) + \arctan \left( \frac{1}{2(k+1)^2} \right)$$

Any advice on how to move forward with this inductive proof? Is there some kind of identity I can use?

Edit

Using the formula that Gary suggested we have \begin{align} \arctan \left( \frac{\frac{k}{k+1} + \frac{1}{2(k+1)^2}}{1 - \left( \frac{k}{k+1} \right) \left( \frac{1}{2(k+1)^2} \right)} \right) &= \arctan \left( \frac{\frac{2k(k+1) + 1}{2(k+1)^2}}{\frac{2(k+1)^2 - k}{2(k+1)^3}} \right) \\ &= \arctan \left( \frac{(2k^2 + 2k + 1)(k+1)}{2(k^3 + 3k^2 + 3k + 1) - k} \right) \\ &= \arctan \left( \frac{(2k^2 + 2k + 1)(k+1)}{2k^3 + 6k^2 + 5k + 2} \right) \\ &= \arctan \left( \frac{(2k^2 + 2k + 1)(k+1)}{(2k^2 + 2k + 1)(k+2)} \right) \\ &= \arctan \left( \frac{k+1}{k+2} \right) \end{align}

Answer 1

I'm stuck on what to do with $$\arctan \left( \frac{k}{k+1} \right) + \arctan \left( \frac{1}{2(k+1)^2} \right)$$

Using one of the identities in the document that the original poster cited,

$$\tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)}. \tag1 $$

Also, as has been indicated in the comments, for any real number $~r,~$ in order for $~\theta~$ to equal $~\arctan(r),~$ two things have to happen:

• You must have that $~\tan(\theta) = r.$

• You must have that $~-\pi/2 < \theta < \pi/2.~$
  Alternatively, in degrees, you must have that $~-90^\circ < \theta < 90^\circ.$

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Continuing where the original poster was stuck:

Let $~a = \arctan\left( ~\dfrac{k}{k+1} ~\right).$

Let $~b = \arctan\left( ~\dfrac{1}{2(k+1)^2} ~\right).$

This implies that $~-\pi/2 < a,b < \pi/2.$

Let $~c = \arctan\left( ~\dfrac{k+1}{k+2} ~\right).$

In order to complete the problem, you have to show that:

• $~\tan(a + b) = \tan(c).$

• $-\pi/2 < (a+b) < \pi/2.$

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$$\tan(a) + \tan(b) \\ = \frac{[ ~2k(k+1)^2 ~] + (k+1) ~]}{2(k+1)^3} = \frac{(k+1) \times [ ~(2k)(k+1) + 1 ~]}{2(k+1)^3} \\ = \frac{(k+1) \times (2k^2 + 2k + 1)}{2(k+1)^3}.$$

Note that

$$(k+2) \times (2k^2 + 2k + 1) = 2(k+1)^3 - k.$$

Then

$$1 - \tan(a)\tan(b) = 1 - \left[ ~\frac{k}{k+1} \times \frac{1}{2(k+1)^2} ~\right] \\ = \frac{2(k+1)^3 - k}{2(k+1)^3} = \frac{(k+2) \times (2k^2 + 2k + 1)}{2(k+1)^3}.$$

Therefore,

$$\tan(a+b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} \\ = \frac{(k+1) \times (2k^2 + 2k + 1)}{(k+2) \times (2k^2 + 2k + 1)} = \frac{k+1}{k+2} = \tan(c).$$

Furthermore, since $~0 < \tan(a),\tan(b) < 1,~$ you have that $0 < a,b < \pi/4 \implies 0 < (a+b) < \pi/2.$

Thus, $(a+b)$ is the unique angle between $~0~$ and $~\pi/2~$ such that $~\tan(a+b) = \tan(c).~$

Therefore,

$$~(a+b) = \arctan\left( ~\frac{k+1}{k+2} ~\right).$$

Therefore,

$$\arctan\left( ~\frac{k}{k+1} ~\right) + \arctan\left( ~\frac{1}{2(k+1)^2} ~\right) = \arctan\left( ~\frac{k+1}{k+2} ~\right).$$

Answer 2

Telescoping may be faster

Noting that $$ \begin{aligned} & \tan \left[\arctan \left(\frac{r}{r+1}\right)-\arctan \left(\frac{r-1}{r}\right)\right] = \frac{\frac{r}{r+1}-\frac{r-1}{r}}{1+\frac{r}{r+1} \cdot \frac{r-1}{r}} = \frac{1}{2 r^2}, \end{aligned} $$ we have $$\arctan \left(\frac{1}{2 r^2}\right)=\arctan \left(\frac{r}{r+1}\right)-\arctan\left(\frac{r-1}{r}\right)$$ Summing them up yields $$ \sum_{r=1}^n \arctan \left(\frac{1}{2 r^2}\right) =\sum_{r=1}^n\left[\arctan \left(\frac{r}{r+1}\right)-\arctan \left(\frac{r-1}{r}\right)\right]= \arctan \left(\frac{n}{n+1}\right) $$