0 to infinity Σsin(14/n) - sin(14/(n+1)) = sin14? why?

You have one summation sign, not two. That makes a big difference here.

Each term consists of the difference of two terms.

S1 = first term = sin(14) - sin(14/2).

S2 = sum of first 2 terms = sin(14) - sin(14/2) + sin(14/2) - sin(14/3) = sin(14) - sin(14/3)

S3 = sum of first 3 terms = sin(14) - sin(14/3) + sin(14/3) - sin(14/4) = sin(14) - sin(14/4)

etc

Sn = sum of first n terms = sin(14) - sin(14/(n+1))

lim (n->infinity) Sn = lim(n->infinity) sin(14) - sin(14/(n+1)) = sin(14)

You can't rearrange that one sum into the two separate diverging sums, because rearranging is only valid when the two individual sums converge.

Here's a more obvious (I hope) example.

Suppose a_n = n - n for all n?

Then obviously a_n = 0 for all n, and Sn = sum(k=1,n) a_n = 0, and lim(n->infinity) Sn = 0.

But can't I just rewrite sum(n - n) as sum(n) - sum(n), both of which diverge? Answer: no.

Expand it out

sin(14/1) - sin(14/2) + sin(14/2) - sin(14/3) + sin(14/3) - sin(14/3) + .... - sin(14/(n + 1))

n goes to infinity, so sin(14/(n + 1)) goes to sin(0), which is 0

Now we have:

sin(14/1) + 0 + 0 + 0 + 0 + ... + sin(0) =>

sin(14) + sin(0) =>

sin(14) + 0 =>

sin(14)

Telescoping series. Learn them, love them.

Σ[n=0 to infinity] sin(14/(n+1)) = Σ[n=1 to infinity] sin(14/(n))

Σ[n=0 to infinity] sin(14/(n) - Σ[n=1 to infinity] sin(14/(n))

sin 14 + Σ[n=1 to infinity] sin(14/(n) - Σ[n=1 to infinity] sin(14/(n))

sin 14