All Questions
1
question
47
votes
1
answer
2k
views
Why does this ratio of sums of square roots equal $1+\sqrt2+\sqrt{4+2\sqrt2}=\cot\frac\pi{16}$ for any natural number $n$?
Why is the following function $f(n)$ constant for any natural number $n$?
$$f(n)=\frac{\sum_{k=1}^{n^2+2n}\sqrt{\sqrt{2n+2}+{\sqrt{n+1+\sqrt k}}}}{\sum_{k=1}^{n^2+2n}\sqrt{\sqrt{2n+2}-{\sqrt{n+1+\...
- 145k