Video: Partial Sums of Alternating Series

Calculate the partial sum 𝑆_𝑛 for the least 𝑛 terms which guarantees that the sum of the first 𝑛 terms of the alternating series βˆ‘_(𝑛 = 1) ^(∞) ((βˆ’1)^(𝑛 + 1))/(5^(𝑛)) differs from the infinite sum by 10⁻⁢ at most. Give your answer to 6 decimal places.

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Video Transcript

Calculate the partial sum 𝑆 𝑛 for the least 𝑛 terms which guarantees that the sum of the first 𝑛 terms of the alternating series the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 plus one divided by five to the 𝑛th power differs from the infinite sum by 10 to the power of negative six at most. Give your answer to six decimal places.

The question wants us to approximate this alternating series by using a partial sum. It wants us to use an approximation which uses the least number of terms which guarantees that our difference between our estimate and the infinite sum is at most 10 to the power of negative six. Once we found this value of 𝑛, we need to calculate the partial sum to six decimal places. We see that the series given to us is a geometric series with initial term π‘Ž equal to one-fifth and ratio of successive terms, π‘Ÿ, equal to negative one-fifth. So, in this case, we could actually just calculate the value of the infinite sum. Then, all we would need to do is add more and more terms to our partial sums until we’re within 10 to the power of negative six of the infinite sum. And this would work.

However, we don’t know how many terms we would need to add. This could take hundreds and hundreds of terms for all we know. Instead, we’re going to try and approximate how many terms we would need. To help us estimate the value of 𝑛 we need, we recall the following fact about alternating series. If π‘Ž 𝑛 is a positive and decreasing sequence whose limit as 𝑛 approaches ∞ is equal to zero, then the alternating series the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 plus one times π‘Ž 𝑛 is convergent by the alternating series test β€” we’ll call this equal to 𝑆. Then, we know we can bound the absolute value of 𝑆 minus the 𝑛th partial sum by the first term we left out. It’s less than or equal to π‘Ž 𝑛 plus one. We will set π‘Ž 𝑛 equal to one divided by five to the 𝑛th power.

We see if we set π‘Ž equal to one-fifth and π‘Ÿ also equal to one-fifth, then we can see that our sequence π‘Ž 𝑛 is also a geometric sequence. This actually then tells us all the information we need to know. One-fifth to the 𝑛th power is always positive. And the absolute value of one-fifth is less than one, so it’s decreasing. Finally, we know the limit as 𝑛 approaches ∞ of one divided by five to the 𝑛th power is equal to zero. So, we can use this to approximate the difference between the 𝑛th partial sum and the actual value of our infinite series 𝑆. The absolute value of 𝑆 minus the 𝑛th partial sum is less than or equal to one divided by five to the power of 𝑛 plus one.

Remember, we want this error to be at most 10 to the power of negative six. So, if we chose a value of 𝑛 such that one divided by five to the power of 𝑛 plus one was less than or equal to 10 to the power of negative six, then, for this particular value of 𝑛, the absolute value of 𝑆 minus the 𝑛th partial sum is less than or equal to 10 to the power of negative six. So, this value of 𝑛 is sufficient. However, it’s not necessarily the lowest value of 𝑛. So, let’s find the sufficient value of 𝑛. We want one divided by five to the power of 𝑛 plus one to be less than or equal to 10 to the power of negative six.

Both of these terms are positive. So, we take the reciprocal of both sides of this equation and then flip the inequality. We then take the log base five of both sides of this inequality. Then, we just subtract one from both sides of the inequality. Evaluating this gives us 𝑛 is greater than or equal to 7.6, which is the same as saying 𝑛 is greater than or equal to eight. So, what we have done is we have shown the absolute value of 𝑆 minus the eighth partial sum is less than or equal to one divided by five to the power of eight plus one, which itself is less than or equal to 10 to the power of negative six. In other words, when 𝑛 is equal to eight, our approximation is at most off by 10 to the power of negative six.

But remember, the question wants us to find the least number of terms which has this property. We know that eight works, but we need to check whether seven works. To do this, let’s start by calculating the value of 𝑆. We can do this since we’re calculating the infinite sum of a geometric series where the absolute value of the ratio of successive terms is less than one. This infinite sum is equal to π‘Ž divided by one minus π‘Ÿ. So, 𝑆 is equal to one-fifth divided by one minus negative one-fifth, which we can calculate to give us one-sixth.

We could calculate the value of 𝑆 seven by using the formula for a finite sum of a geometric series. However, we can also just use a calculator, giving us approximately 0.166665. If we then calculate the difference between 𝑆 and our seventh partial sum, we see it’s approximately 1.6 times 10 to the power of negative six. And we see that this is bigger than 10 to the power of negative six. So, this means that 𝑛 equals eight must’ve been the least number of terms which guaranteed this level of accuracy. And we can then calculate 𝑆 eight to six decimal places to be 0.166666.

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