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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 {( - \frac {1}{3})^n} {n} (|error| < 0.0005) $
We only need to add the first five terms of the series to approximate the sum within the allotted error.
Calculus 2 / BC
Chapter 11
Infinite Sequences and Series
Section 5
Alternating Series
Sequences
Series
Harvey Mudd College
Baylor University
Idaho State University
Boston College
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let's start off by showing that the syriza's conversion. So here we would use the alternating Siri's test with B N equals one over end, three to the end. This is clearly positive. The limited bien zero. You could see that this term goes to infinity, so the fraction is a whole ghost zero and being is decreasing. This is clearly true because if you just replace and within plus one, that's true. So this thing converges by the alternating Siri's test, so that takes care of the first part of the first sentence. Now we'd like to know how many terms. So what's its end value off the Siri's do we need so that if we take the partial song, then the air when approximating the actual song is less than this fraction over here. So, using the alternating Siri's estimation Tehran, the absolute value of their is less than being plus one. We're here. This is the air when using in terms. So now we want this to be Weston. That's the This is the given number that was given to us. So we'LL find the end that makes this true. So let's what's right This now that's equivalent to end three the end being bigger than two thousand. And then here you would just do some trial in here until you find the first one. If you look at a five not quite large enough. However, when you could say six, this is what we want. This is bigger than the two thousand. So we want and equal six here or really in this case we want and plus one equals six. So that just means and equals five. So we'LL just add five terms is all that is required, and that's our final answer.
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