The problem I have is a summation from 1 to infinity for (-1)^n/n. I have to find a partial sum that is in 0.001 of the infinite sum, my teacher says the fourth partial sum approximates the infinite sum, which is -7/12. I think to get within 0.001 of the infinite sum, do I have to take 999 terms. However, I don't know how right that is, so a little help would be good.
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.
To approximate the infinite sum ∑((-1)^n/n) from 1 to infinity within 0.001, we can use the concept of partial sums.
Let's start by finding the formula for the nth partial sum, Sn, of the series. The nth partial sum is the sum of the first n terms of the series. In this case, the series is given by ∑((-1)^n/n).
To find Sn, we need to calculate the sum of the terms up to the nth term. In other words, we need to evaluate the expression (-1)^1/1 + (-1)^2/2 + ... + (-1)^n/n.
Now, since the series alternates between positive and negative terms, we can group them together to simplify the calculations. This allows us to write the expression for Sn as follows:
Sn = (-1/1 + 1/2) + (-1/3 + 1/4) + ... + [(-1)^(n-1)/(2n-1) + (-1)^n/2n]
By rearranging the terms, we can rewrite it as:
Sn = 1/2 - 1/4 + 1/6 - 1/8 + ... + (-1)^(n-1)/(2n-1) + (-1)^n/2n
Now, to determine the value of n that would give us an approximate sum within 0.001 of the infinite sum, we can compare successive partial sums to the infinite sum itself.
For example, let's calculate the difference between the fourth partial sum and the infinite sum:
Difference = |S4 - ∑((-1)^n/n)|
Since your teacher mentioned that the fourth partial sum is approximately -7/12, we can substitute that in:
Difference = |(-7/12) - ∑((-1)^n/n)|
To get this difference within 0.001, we want |Difference| ≤ 0.001.
So, you would need to calculate additional terms until you find a partial sum in which the difference between the partial sum and the infinite sum is within 0.001.
You can start by calculating S5, S6, S7, and so on, until you find the first partial sum where |S𝑛 - ∑((-1)^n/n)| ≤ 0.001.