Final answer:
To estimate the value of the series ∑(infinity)ₖ₌₁ (-1)ᵏ/(k+1)!, we can use the alternating series estimation theorem. This theorem states that if an alternating series satisfies the conditions of being decreasing in magnitude and having its terms approach zero as k approaches infinity, then the absolute value of the error in the partial sum is less than or equal to the absolute value of the next term. The estimated value of the series is 0.00833 with an absolute error less than 10^(-3).
Step-by-step explanation:
To estimate the value of the series ∑(infinity)ₖ₌₁ (-1)ᵏ/(k+1)!, we can use the alternating series estimation theorem. This theorem states that if an alternating series satisfies the conditions of being decreasing in magnitude and having its terms approach zero as k approaches infinity, then the absolute value of the error in the partial sum is less than or equal to the absolute value of the next term.
In this case, the next term is (-1)^(k+1)/(k+2)!. Since we want the absolute error to be less than 10^(-3), we need to find the smallest value of k for which |(-1)^(k+1)/(k+2)!| < 10^(-3).
We can start by checking if the first term satisfies this condition. Plugging in k = 1, we have |(-1)^(1+1)/(1+2)!| = 1/3! = 1/6 ≈ 0.167, which is greater than 10^(-3). So we continue checking the next terms until we find one that satisfies the condition. The smallest value of k that satisfies the condition is k = 4, which gives |(-1)^(4+1)/(4+2)!| = 1/6! = 1/720 ≈ 0.0014, which is less than 10^(-3).
The estimated value of the series is the sum of the terms up to k = 4. Plugging in k = 4, we have (-1)^(4)/(4+1)! = 1/5! = 1/120 = 0.00833. The absolute error in this partial sum is the absolute value of the next term, which is |(-1)^(4+1)/(4+2)!| = 1/6! = 1/720 ≈ 0.0014. Since this absolute error is less than 10^(-3), the estimated value of the series is 0.00833.