What is the summation of sin(6/n)-sin(6/(n+1)) form 1 to INF?

Answer 1

Lets compute a few terms:

sin(6/1) - sin(6/2)
sin(6/2) - sin(6/3)
sin(6/3) - sin(6/4)

If we add these up note that the first sin() cancels the sin() of the previous term in the sequence. So for example, sin(6/3) in the third term cancels sin(6/3) in the second term.
So when we add up the series all the terms cancel, except for the sin(6/1) in the first term and sin(6/(n+1)) in the last term. As n becomes very large, sin(6/(n+1)) becomes very small.

So the limit of the sum is just sin(6) or approximately -0.23

Answer 2

write it down
sin(6/2)-sin6/3+
sin6/3-sin6/4+ at the end the sum is


sin 6/2-sin(6/n+1) ===> sin 3 as 6/(n+1)==>0