# Taylor Series Remainder Theorem

1. Prove that the MacLaurin series for cosx converges to cosx for all x.

## Homework Equations

Ʃ(n=0 to infinity) ((-1)^n)(x^2n)/((2n)!) is the MacLaurin series for cosx
|Rn(x)|$\leq$M*(|x|^(n+1))/((n+1)!) if |f^(n+1)(x)|$\leq$M
lim(n->infinity)Rn=0 then a function is equal to its Taylor series within the interval of convergence.

3. The Attempt at a Solution . I already know what the solution is but I can't figure out why we are using this solution.
First you get: |f^(n+1)(x)|$\leq$1 because |cosx| and |sinx| functions are always bounded by 1
Then you write |Rn(x)|$\leq$(|x|^(n+1))/((n+1)!)
lim as n->infinity of (|x|^(n+1))/((n+1)!)=0, therefore lim as n-> infinity of |Rn|=0 so the MacLaurin series of cosx converges to cosx for all x.
My question
Why don't we just write Rn = $\Sigma$ from (i=n+1 to infinity) ((-1)^i)*(x^2i)/((2i)!) and we can see from this that the limit as n->infinity makes Rn go to zero, therefore cosx converges to cosx? Why did we have to go through all of that other stuff?