How to find the empirical formula?

How to find a sequence formula from a series formula?

• How do you find a sequence formula An (a sub n) from a given series formula Sn (s sub n? We know that the limit as n approaching infinity of Sn is equal to the summation to infinity of the sequence An. The formula I have been given is An = Sn - Sn-1 (Sn-1 is s sub (n-1)) The basic concept is that Sn = A1 + A2 + A3 + A4 +...+An and Sn-1 = A0 + A1 + A2 + A3 +....+ An-1 All this cancels out except for the An and the A0, so this formula works if A0 = 0 as in An = n/(n+1), but the problem I am trying to find out has a value A0 that doesn't equal zero, so this formula doesn't work. I am wondering if there is an assumption made here that makes the formula work if you just make that an assumption or if there is a flaw in my math or just in the formula itself. The problem that brought this up is: Given Sn = 3+ n2^(-n), find An and the summation to infinity of An with an initial value n=1. The summation to An will be the limit as n approaches infinity of Sn, so that is easy. Use L'Hospital's Rule and you end up with 3. What I obtained for the formula above, however cancels out the three, making it completely inconsequential. I plugged the formula into a sum on Wolfram Alpha, and it gave me that it went to 0. Is my math wrong, or is the formula not being used correctly, or what?

• Answer:

  This is correct; you just need to remember to initialize S(1) (or whatever the first index is) as A(1). For instance, A(1) = S(1) = 3 + 1 * 2^(-1) = 5/2, and for all n > 1, A(n) = S(n) - S(n-1) .......= [3 + n * 2^(-n)] - [3 + (n-1) * 2^(-(n-1))] .......= n * 2^(-n) - (n-1) * 2^(-n+1) .......= n * 2^(-n) - 2(n-1) * 2^(-n) .......= [n - 2(n-1)] 2^(-n) .......= (2 - n) * 2^(-n). Note: This latter formula for A(n) works for n > 1. I hope this helps!