Posted by Sophie on Tuesday, May 5, 2009 at 12:21am.
The infinite sum from n = 1 to infinity is not -7/12.
-1 +1/2 -1/3 +1/4 = -7/12 = -0.58333, but the infinite sum is -ln2 = -0.69315...
You will have to add a very large number of terms to get within 0.001, since the 1000th term will be 0.001 and there will be smaller fluctuations about the limit after that.
This is how you prove it:
In case of a alternating summation where the absolute values of the terms descrease monotonically, the partial summations are always upper or lower bounds to the infinite summation. To see this write the infinite summation S as:
S = S_N + Tail(N+1)
Where S_N is the partial sum of the first N terms and the Tail(k) is the summation from n = k to infinity.
Write the n-th term in the summation as
(-1)^n a_n
with a_n positive (in our case a_n = 1/n).
We assume that a_{n+1} < a_{n} which is clearly true in the present case.
The tail of the summation can be written as:
Tail(N+1) = Limit M to infinity of
(-1)^(N+1) [a_{N+1} - a_{N+2} +
a_{N+3} - a_{N+4} + ...+(-1)^(M+1)a_{M}]
If you group together each two consecutive terms in the square brackets, you see that every negative term is paired up with a positive term and you may have one extra positive term (if M is odd), so whatever the value of the square bracket it, you know that it is larger than zero.
So, we have that:
Tail(N+1) = (-1)^(N+1) Limit M to infinity of f(M)
where f(M) > 0.
It then follows that Tail(N+1) has a sign of (-1)^(N+1)
So, for even N, Tail(N+1) will be negative, meaning that S_N will be larger than the limit, while for odd N Tail(N+1) will be positive and thus S_N will be lower than the limit.
We thus have:
S_{999} < S
S_{1000} > S
The difference between S_{1000} and
S_{999} is thus larger than the difference between the difference between S and S_{999}. So, the error when you include only the first 999 terms will be less than
S_{1000} - S_{999} = 1/000.