Base Case: n = 1

We begin by verifying the statement for n = 1 .

\sum_{r=1}^{1}\arctan\left(\frac{2}{r^2}\right) = \arctan\left(\frac{2}{1^2}\right) = \arctan(2)

On the right-hand side of the equation:

\arctan(1) + \arctan(2) - \frac{\pi}{4}

We know that:

\arctan(1) = \frac{\pi}{4}

Substituting this back:

\frac{\pi}{4} + \arctan(2) - \frac{\pi}{4} = \arctan(2)

Both sides are equal, so the base case holds .

Assume true for n=k

Assume that the statement is true for some n = k , where k \in \mathbb{N} . That is,

\sum_{r=1}^{k}\arctan\left(\frac{2}{r^2}\right) = \arctan(k) + \arctan(k+1) - \frac{\pi}{4}

Induction Step: Prove for n = k + 1

We need to show that:

\sum_{r=1}^{k+1}\arctan\left(\frac{2}{r^2}\right) = \arctan(k+1) + \arctan(k+2) - \frac{\pi}{4}

Starting with the left-hand side:

\sum_{r=1}^{k+1}\arctan\left(\frac{2}{r^2}\right) = \left( \sum_{r=1}^{k}\arctan\left(\frac{2}{r^2}\right) \right) + \arctan\left(\frac{2}{(k+1)^2}\right)

Applying the induction hypothesis :

= \arctan(k) + \arctan(k+1) - \frac{\pi}{4} + \arctan\left(\frac{2}{(k+1)^2}\right)

From part (d), we have:

\arctan(a+1)-\arctan(a-1)=\arctan\left(\frac{2}{a^2}\right)

Using this identity for a=k+1 :

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

Substituting back:

\sum_{r=1}^{k+1}\arctan\left(\frac{2}{r^2}\right) = \arctan(k) + \arctan(k+1) - \frac{\pi}{4} + \arctan(k+2) - \arctan(k)

Simplifying:

= \arctan(k+1) + \arctan(k+2) - \frac{\pi}{4}

This matches the right-hand side of the statement for n = k + 1 .

Conclusion

We have shown that the statement is true for n=1, and that it if it is true for n=k, then is true for n=k+1. By the principle of mathematical induction, the statement

\sum_{r=1}^{n}\arctan\left(\frac{2}{r^2}\right)=\arctan(n)+\arctan(n+1)-\frac{\pi}{4}

holds for all n \in \mathbb{N} .

Problem Bank - Proof and Reasoning

Problem Bank - Proof and Reasoning