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$\arctan\left(\frac{1}{2\left(1\right)^{2}}\right)+\arctan\left(\frac{1}{2\left(2\right)^{2}}\right)+\arctan\left(\frac{1}{2\left(3\right)^{2}}\right)+...+\arctan\left(\frac{1}{2\left(n\right)^{2}}\right)=\frac{\pi}{4}-\arctan\left(\frac{1}{2n+1}\right)$

I assume you need to do something like "take tan of both sides" but that doesn't really work for me given the summation on the LHS. Any hints?

user71207
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    Maybe this helps: $\pi/4-arctan(\frac{1}{2n+1})+arctan(\frac{1}{2{(n+1)^2}})=\pi/4-arctan(\frac{1}{2n+3})$ – imranfat Nov 21 '20 at 03:34
  • Hint use $\arctan(x)+\arctan(y)$ formula incrementally and show you get $\frac 12,\frac 23,\frac 34,\frac 45,\cdots$ sequence as inners of artcan. What is the term when you sum them all. Move the RHS arctan on the LHS and apply the formula once more to get $\pi/4$. – zwim Nov 21 '20 at 03:34
  • @imranfat yeah so basically I need to prove $\left(\frac{1}{2n+1}\right)+\left(\frac{1}{2\left(n+1\right)^{2}}\right)=\left(\frac{1}{2n+3}\right)$

    But I'm not sure which method to use. How can I combine the two on the LHS into one fraction?

     – user71207 Nov 22 '20 at 06:46

1 Answers1

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I'll assume you were able to prove the base case $n = 1$.

The inductive hypothesis is to assume $$\arctan \dfrac{1}{2\cdot1^2}+\cdots+\arctan \dfrac{1}{2\cdot n^2} = \dfrac{\pi}{4} - \arctan \dfrac{1}{2n+1}. \quad (1)$$

Then, you need to prove $$\arctan \dfrac{1}{2\cdot1^2}+\cdots+\arctan \dfrac{1}{2\cdot n^2} +\arctan \dfrac{1}{2\cdot (n+1)^2} = \dfrac{\pi}{4} - \arctan \dfrac{1}{2n+3}. \quad (2)$$

If you subtract $(1)$ from $(2)$, you get the equation $$\arctan \dfrac{1}{2\cdot (n+1)^2} = \arctan \dfrac{1}{2n+1} - \arctan \dfrac{1}{2n+3}. \quad (3)$$

So, after assuming the inductive hypothesis $(1)$, you need to prove equation $(3)$. Then you can add equation $(3)$ to equation $(1)$ to prove equation $(2)$, which completes the inductive step.

JimmyK4542
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