Respuesta :
Answer:
The answer is "4 terms".
Step-by-step explanation:
[tex]\sum_{n=1}^{\infty} \frac{(-1)^{n+2} (0.4)^{2n+1}}{(2n+1)!} \ \ \ \ \ error \leq 0.0000001\\\\\\\sum_{n=1}^{\infty} \frac{(-1)^{n+2} (0.4)^{2n+1}}{(2n+1)!} = \frac{-1^{0+2}}{1!}(0.4) + \frac{-1^{1+2}}{3!}(0.4)^3 +\frac{-1^{2+2}}{5!}(0.4)^5+\frac{-1^{3+2}}{7!}(0.4)^7+..\\\\[/tex]
[tex]=0.4 - \frac{0.4^3}{3!} + \frac{0.4^5}{5!} - \frac{0.4^7}{7!}+ \frac{0.4^9}{9!} - \frac{0.4^{11}}{11!}+............\\[/tex]
Note that, alternatively, positive and negative terms are given in the sequence. It is also alternating
Apply alternative test series:
[tex]\sum_{n=1}^{\infty} (-1)^{n-1} b_n = b_1-b_2+b_3-b_4+b_5-......+(-1)^{n-1} b_n >0\\\\i) b_{n+1}<b_n \ \ \ \ \forall_n \\\\ii) lim_{x\longrightarrow \infity} \ b_n = 0[/tex]
use alternating test method we get:
[tex]\sum_{n=1}^{\infty} \frac{-1^{n-1}}{n^2}\\\\s =0.4 - \frac{0.4^3}{3!} + \frac{0.4^5}{5!} - \frac{0.4^7}{7!}+ \frac{0.4^9}{9!} - \frac{0.4^{11}}{11!}+..\\\\b_5= \frac{0.4^9}{9!} = 0.00000000072 < \frac{1}{10^7} =0.0000001[/tex]
that's why its value is 4 terms
The approximation 0.0000001 within the convergent value of that series the fourth term.
What is an alternative test series?
It is the convergent series when it is term decrease in absolute value and approaches the zero limit.
[tex]\sum_{n=1}^{\infty} \dfrac{(-1)^{n+2}(0.4)^{2n+1}}{(2n+1)!}[/tex]
When error ≤ 0.0000001
[tex]\begin{aligned} \sum_{n=1}^{\infty} \dfrac{(-1)^{n+2}(0.4)^{2n+1}}{(2n+1)!} &= \dfrac{(-1)^{0+2}}{1!}(0.4) + \dfrac{(-1)^{1+2}}{3!}(0.4)^{3} + ..............\\\\&= 0.4 - \dfrac{0.4^{3}}{3!} + \dfrac{0.4^{5}}{5!} - \dfrac{0.4^{7}}{7!} + ...............\end{aligned}[/tex]
Now by the alternative test series, we have
[tex]\Sigma _{n=1}^{\infty} (-1)^{n-1} b_n =b_1-b_2+b_3-b_4+....+(-1)^{n-1}b_n > 0\\[/tex]
Then
[tex]i) \ \ b_{n+1} < b_n \ \ \ \ \forall _n\\\\ii) \ \ lim_{x \rightarrow } b_n = 0[/tex]
Then by the alternative test method, we have
[tex]\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n^{2}} \\\\s = 0.4 - \dfrac{0.4^3}{3!} + \dfrac{0.4^5}{5!} -\dfrac{0.4^7}{7!}+....\\\\b_5 = \dfrac{0.4^9}{9!} = 0.00000000072 < \dfrac{1}{10^7} = 0.0000001[/tex]
More about the alternative test series link is given below.
https://brainly.com/question/16969349
