Question

how many terms do you have to compute in order for your approximation (your partial sum) to be within 0.0000001 from the convergent value of that series

Answer 1

Answer:

The answer is "4 terms".

Step-by-step explanation:

$\sum_{n=1}^{\infty} \frac{(-1)^{n+2} (0.4)^{2n+1}}{(2n+1)!} \ \ \ \ \ error \leq 0.0000001\\\\\\\sum_{n=1}^{\infty} \frac{(-1)^{n+2} (0.4)^{2n+1}}{(2n+1)!} = \frac{-1^{0+2}}{1!}(0.4) + \frac{-1^{1+2}}{3!}(0.4)^3 +\frac{-1^{2+2}}{5!}(0.4)^5+\frac{-1^{3+2}}{7!}(0.4)^7+..$

                                  $=0.4 - \frac{0.4^3}{3!} + \frac{0.4^5}{5!} - \frac{0.4^7}{7!}+ \frac{0.4^9}{9!} - \frac{0.4^{11}}{11!}+............$

Note that, alternatively, positive and negative terms are given in the sequence. It is also alternating  

Apply alternative test series:

$\sum_{n=1}^{\infty} (-1)^{n-1} b_n = b_1-b_2+b_3-b_4+b_5-......+(-1)^{n-1} b_n >0\\\\i) b_{n+1}$

use alternating test method we get:

$\sum_{n=1}^{\infty} \frac{-1^{n-1}}{n^2}\\\\s =0.4 - \frac{0.4^3}{3!} + \frac{0.4^5}{5!} - \frac{0.4^7}{7!}+ \frac{0.4^9}{9!} - \frac{0.4^{11}}{11!}+..\\\\b_5= \frac{0.4^9}{9!} = 0.00000000072 < \frac{1}{10^7} =0.0000001$

that's why its value is 4 terms