Answer :
To show the convergence of the series
[tex]\[ \sum_{n=0}^{\infty} \frac{(-1)^n}{5^n n!}, \][/tex]
we first need to understand the nature of this series. This series is related to the Maclaurin series expansion for the exponential function [tex]\(e^x\)[/tex].
Consider the exponential function [tex]\(e^{-x}\)[/tex], whose Maclaurin series expansion is given by:
[tex]\[ e^{-x} = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!}. \][/tex]
By substituting [tex]\(x = \frac{1}{5}\)[/tex] into the series, we get:
[tex]\[ e^{-\frac{1}{5}} = \sum_{n=0}^{\infty} \frac{(-\frac{1}{5})^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n}{5^n n!}. \][/tex]
Since the exponential function [tex]\(e^{-x}\)[/tex] is known to be convergent for any real [tex]\(x\)[/tex], the series
[tex]\[ \sum_{n=0}^{\infty} \frac{(-1)^n}{5^n n!} \][/tex]
is convergent.
Next, we need to determine how many terms of the series we need to add to find the sum to the indicated accuracy [tex]\( \mid \text {error} \mid < 0.000005 \)[/tex].
We can start by calculating the terms of the series until the absolute value of the next term is smaller than the desired accuracy [tex]\(0.000005\)[/tex].
The general term of the series is:
[tex]\[ a_n = \frac{(-1)^n}{5^n n!} \][/tex]
To find when the terms are sufficiently small, we will sum the series terms and check their magnitude.
Here are the first few terms of the series and their sum:
[tex]\[ \begin{aligned} a_0 & = \frac{(-1)^0}{5^0 \cdot 0!} = 1, \\ a_1 & = \frac{(-1)^1}{5^1 \cdot 1!} = -\frac{1}{5}, \\ a_2 & = \frac{(-1)^2}{5^2 \cdot 2!} = \frac{1}{50}, \\ a_3 & = \frac{(-1)^3}{5^3 \cdot 3!} = -\frac{1}{750}, \\ a_4 & = \frac{(-1)^4}{5^4 \cdot 4!} = \frac{1}{12500}, \\ a_5 & = \frac{(-1)^5}{5^5 \cdot 5!} = -\frac{1}{312500}. \end{aligned} \][/tex]
Summing these terms until the magnitude of the terms is less than the desired accuracy:
[tex]\[ \begin{aligned} \text{Sum} &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 \\ &= 1 - \frac{1}{5} + \frac{1}{50} - \frac{1}{750} + \frac{1}{12500} - \frac{1}{312500} \\ &\approx 0.8187306666666667. \end{aligned} \][/tex]
To ensure [tex]\( \mid \text {error} \mid < 0.000005 \)[/tex], we check the next term [tex]\(a_6\)[/tex]:
[tex]\[ a_6 = \frac{(-1)^6}{5^6 \cdot 6!} = \frac{1}{15625000} \approx 2.666666666666667 \times 10^{-6}. \][/tex]
The term [tex]\(a_6\)[/tex] is indeed smaller than 0.000005, confirming that the approximation meets the required accuracy.
Thus, we need to add 6 terms of the series to achieve the indicated accuracy, so the number of terms required is [tex]\(6\)[/tex].
[tex]\[ \sum_{n=0}^{\infty} \frac{(-1)^n}{5^n n!}, \][/tex]
we first need to understand the nature of this series. This series is related to the Maclaurin series expansion for the exponential function [tex]\(e^x\)[/tex].
Consider the exponential function [tex]\(e^{-x}\)[/tex], whose Maclaurin series expansion is given by:
[tex]\[ e^{-x} = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!}. \][/tex]
By substituting [tex]\(x = \frac{1}{5}\)[/tex] into the series, we get:
[tex]\[ e^{-\frac{1}{5}} = \sum_{n=0}^{\infty} \frac{(-\frac{1}{5})^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n}{5^n n!}. \][/tex]
Since the exponential function [tex]\(e^{-x}\)[/tex] is known to be convergent for any real [tex]\(x\)[/tex], the series
[tex]\[ \sum_{n=0}^{\infty} \frac{(-1)^n}{5^n n!} \][/tex]
is convergent.
Next, we need to determine how many terms of the series we need to add to find the sum to the indicated accuracy [tex]\( \mid \text {error} \mid < 0.000005 \)[/tex].
We can start by calculating the terms of the series until the absolute value of the next term is smaller than the desired accuracy [tex]\(0.000005\)[/tex].
The general term of the series is:
[tex]\[ a_n = \frac{(-1)^n}{5^n n!} \][/tex]
To find when the terms are sufficiently small, we will sum the series terms and check their magnitude.
Here are the first few terms of the series and their sum:
[tex]\[ \begin{aligned} a_0 & = \frac{(-1)^0}{5^0 \cdot 0!} = 1, \\ a_1 & = \frac{(-1)^1}{5^1 \cdot 1!} = -\frac{1}{5}, \\ a_2 & = \frac{(-1)^2}{5^2 \cdot 2!} = \frac{1}{50}, \\ a_3 & = \frac{(-1)^3}{5^3 \cdot 3!} = -\frac{1}{750}, \\ a_4 & = \frac{(-1)^4}{5^4 \cdot 4!} = \frac{1}{12500}, \\ a_5 & = \frac{(-1)^5}{5^5 \cdot 5!} = -\frac{1}{312500}. \end{aligned} \][/tex]
Summing these terms until the magnitude of the terms is less than the desired accuracy:
[tex]\[ \begin{aligned} \text{Sum} &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 \\ &= 1 - \frac{1}{5} + \frac{1}{50} - \frac{1}{750} + \frac{1}{12500} - \frac{1}{312500} \\ &\approx 0.8187306666666667. \end{aligned} \][/tex]
To ensure [tex]\( \mid \text {error} \mid < 0.000005 \)[/tex], we check the next term [tex]\(a_6\)[/tex]:
[tex]\[ a_6 = \frac{(-1)^6}{5^6 \cdot 6!} = \frac{1}{15625000} \approx 2.666666666666667 \times 10^{-6}. \][/tex]
The term [tex]\(a_6\)[/tex] is indeed smaller than 0.000005, confirming that the approximation meets the required accuracy.
Thus, we need to add 6 terms of the series to achieve the indicated accuracy, so the number of terms required is [tex]\(6\)[/tex].
