Question

Use the Alternating Series Approximation Theorem to find the sum of the series sigma^infinity_n = 1 (-1)^n - 1/n! with less than 0001 error.

Answer 1

Answer:

$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198 -0.0000248$

For the 7th term we have 3 decimals of approximation but our value is 0.000198 higher than the error required, so we can use the 8th term and we have that $|-0.0000248|= 0.0000248$ and with this we have 4 decimals of approximation so if we add the first 8 terms we have a good approximation for the series with an error bound lower than 0.0001.

$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198-0.0000248 =0.632118$

Step-by-step explanation:

Assuming the following series:

$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!}$

We want to approximate the value for the series with less than 0.0001 of error.

First we need to ensure that the series converges. If we have a series $\sum a_n$ where $a_n = (-1)^n b_n$ [/tex] or $a_n =(-1)^{n-1} b_n$ where $b_n \geq 0$ for all n if we satisfy the two conditions given:

1) $lim_{n \to \infty} b_n =0$

2) {$b_n$} is a decreasing sequence

Then $\sum a_n$ is convergent. For this case we have that:

$lim_{n \to \infty} \frac{1}{n!} =0$

And $\frac{1}{n!} <\frac{1}{(n-1)!}$ because $\frac{1}{n!} =\frac{1}{n (n-1)!}$ and $\frac{1}{n(n-1)!} < \frac{1}{(n-1)!}$

So then we satisfy both conditions and then the series converges. Now in order to find the approximation with the error required we can write the first terms for the series like this:

$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198 -0.0000248$

For the 7th term we have 3 decimals of approximation but our value is 0.000198 higher than the error required, so we can use the 8th term and we have that $|-0.0000248|= 0.0000248$ and with this we have 4 decimals of approximation so if we add the first 8 terms we have a good approximation for the series with an error bound lower than 0.0001.

$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198-0.0000248 =0.632118$