The series the summation from n equals 1 to infinity of the quotient of negative 1 raised to the n plus 1 power and n is convergent. Use the Alternating Series Test to find an upper bound on the absolute error if the 7th partial sum is used as an estimate of the infinite sum.
Type your answer in the space below and give three decimal places. If your answer is less than 1, place a leading "0" before the decimal point (ex: 0.482). (10 points)
To use the Alternating Series Test, we first need to verify that the terms of the series are decreasing in absolute value.
The terms of the series are given by \((-1)^{n+1}/n\).
Let's compare the nth term, \(a_n\), with the (n+1)th term, \(a_{n+1}\), to see if they are decreasing:
\[
|a_n| - |a_{n+1}| = \left|\frac{{(-1)^{n+1}}}{n}\right| - \left|\frac{{(-1)^{n+2}}}{n+1}\right|
\]
Next, let's simplify this expression:
\[
\left|\frac{{(-1)^{n+1}}}{n}\right| - \left|\frac{{(-1)^{n+2}}}{n+1}\right| = \frac{1}{n} - \frac{1}{n+1}
\]
Combining the fractions gives us:
\[
\frac{1}{n} - \frac{1}{n+1} = \frac{n+1 - n}{n(n+1)} = \frac{1}{n(n+1)}
\]
Since \(\frac{1}{n(n+1)}\) is positive, we can conclude that the terms of the series are indeed decreasing in absolute value.
Now, let's find an upper bound on the absolute error if the 7th partial sum is used as an estimate of the infinite sum.
The formula for the error approximation using the Alternating Series Test is:
\[
|E_n| \leq |a_{n+1}|
\]
where \(E_n\) represents the absolute error for the nth partial sum.
In this case, we want the absolute error for the 7th partial sum:
\[
|E_7| \leq |a_8|
\]
Substituting the value of \(a_{n+1}\) into the equation gives us:
\[
|E_7| \leq \left|\frac{{(-1)^{8+1}}}{8}\right| = \left|\frac{1}{8}\right|
\]
Therefore, the upper bound on the absolute error if the 7th partial sum is used as an estimate of the infinite sum is \(0.125\).
To determine if the series is convergent using the Alternating Series Test, we need to check two conditions:
1. The terms of the series must alternate sign.
2. The absolute value of each term must approach zero as n approaches infinity.
Looking at the given series:
-1/1, 1/2, -1/3, 1/4, -1/5, 1/6, ...
1. The terms alternate sign. Each term alternates between negative and positive, as required by the Alternating Series Test.
2. The absolute value of each term approaches zero. As n gets larger, the terms become smaller and closer to zero.
Therefore, the given series satisfies the conditions of the Alternating Series Test and is convergent.
To find an upper bound on the absolute error, we need to estimate the difference between the 7th partial sum and the actual infinite sum.
The 7th partial sum can be calculated by summing the first 7 terms of the series:
S7 = -1/1 + 1/2 - 1/3 + 1/4 - 1/5 + 1/6 - 1/7
To estimate the difference between S7 and the infinite sum, we use the formula for the absolute error of an alternating series:
Error ≤ Absolute Value of the (n+1)st term
In this case, the (n+1)st term is the 8th term. Let's calculate it:
8th term = -1/8
Now, we can find the absolute error:
Error ≤ Absolute Value of (-1/8)
Absolute error = 1/8
Finally, we need to express the absolute error as a decimal up to three decimal places:
Absolute error = 0.125
Therefore, the upper bound on the absolute error, when using the 7th partial sum as an estimate of the infinite sum, is 0.125.
Recall that for an alternating series, the error is bounded by the first term discarded. That is
|S-S_n| < a_(n+1)