As a little update on my most recent proof, I spent the last half hour of so on my statistics exam generalising the proof further. Instead of using 2
x, you can use any function that diverges to infinity as x goes to infinity. And to further generalise, given that lim x→infinity of g(x) goes to infinity, then lim x→infinity of g(x)sin(k/g(x))=k. However, using g(x)=2
x reveals the neat identity with the nested radicals
Also, g(x)tan(k/g(x))=k as x→infinity. It converges to k from above, whereas the sine version converges from below. The reason is simple : g(x)tan(k/g(x))=g(x)sin(k/g(x))/cos(k/g(x)). As x→infinty, this converges to g(x)sin(k/g(x))/1=g(x)sin(k/g(x))
I also created a really complicated looking problem that evaluated to the partial sum of k
2, too, just for fun. I used:
lim a→inf sum(e
2k+1sin
2(k*e
-2k),k,a,n+a)
Hehe ^_^
EDIT: In the last equation, I had to replace the 'x' with 'k'
Also, a prettier image is attached.
EDIT2: In the picture, pretend I put the
2 symbol for sin
2() >.>
