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Numerade Educator



Problem 25 Easy Difficulty

Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy?

$ \displaystyle \sum_{n = 1}^{\infty} \frac {( - 1)^{n-1}}{n^2 2^n} (|error| < 0.0005) $


That is, since the 6 th term is less than the desired error, we need to add the first 5 terms to get the sum to the desired accuracy.


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Video Transcript

Let's show that the syriza's conversion since it's alternating Siri's with continues the alternating Siri's test. So this requires that first, we define being to just be the positive par of the term. And there's three conditions that have to be satisfied. The first one is that beings positive. That's true. The next one is that the limit of D n A as and goes to infinity a zero. That's true. If you take the limit of this term up here, the numerator is just one, but the denominator goes to infinity so that fraction will go to zero, and the last condition is that the beyond sequence is decreasing. To see that this is true. Just go ahead and replace and with n Plus One, and this is definitely less than bien. The reason this is true is because the denominator on the left is larger than the denominator on the right. The larger the denominator, the smaller the freshen. So therefore, the Siri's will converge by the alternating Siri's test. So that's the first part of this question now, since we showed it converges by the alternating Siri's test, we should go ahead and use the alternating Siri's estimation there, Um, also in eleven point five this section. This tells us that the ear an absolute value one of using in terms to approximate the infinite sum is less than or equal being plus one. And we want that to be less than zero point zero zero zero five. Of course, the next cage. So we want if we want to solve this for n so equivalently two to the end and squared is bigger than two thousand. And here the first time this is true is when and his larger six or more. However we wanted we wanted to be in plus one to be less than zero point zero zero zero five. So, really, in our case, we should have and plus one bigger than equal to six. So that means and bigger than or equal to five. So we should choose five terms, and that's our final answer.