Video: Find the Lowest Bound on the Number of Terms to Guarantee a Level of Accuracy in an Alternating Series

Find the lowest 𝑛 that guarantees that the partial sum 𝑆_(𝑛) of the series 𝑆 = βˆ‘_(𝑛 = 1)^(∞) (βˆ’1)^(𝑛) (3/(6^(𝑛) + 10)) differs from the infinite sum by 10⁻⁸ at the most.

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Video Transcript

Find the lowest 𝑛 that guarantees that the partial sum 𝑆 𝑛 of the series 𝑆 is equal to the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 times three divided by six to the 𝑛th power plus 10 differs from the infinite sum by 10 to the power of negative eight at the most.

The question gives us an infinite series, which converges to some value 𝑆. The question wants us to approximate this value of 𝑆 by taking a partial sum, which means approximating 𝑆 by taking the sum of a finite number of terms of this series. We need to find the lowest number of terms, 𝑛, which guarantees that our approximation differs from the actual value of 𝑆 by at most 10 to the power of negative eight. In other words, the absolute value of 𝑆 minus the 𝑛th partial sum needs to be less than or equal to 10 to the power of negative eight.

And since we want the lowest number of terms, if we were to take one less term, we need this error to be bigger than 10 to the power of negative eight. We see the series given to us in the question appears to be an alternating series. And we’ve seen bounds of this form on certain alternating series before. We recall if we have a positive and decreasing sequence π‘Ž 𝑛 where the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 is equal to zero. Then by the alternating series test the sum from 𝑛 equals one to ∞ of negative one to the 𝑛th power times π‘Ž 𝑛 converges. We’ll say this is equal to 𝑆.

And for these types of alternating series, we know we can approximate the value of 𝑆 by taking the 𝑛th partial sum. In particular, we know the absolute value of 𝑆 minus the 𝑛th partial sum is less than or equal to π‘Ž 𝑛 plus one. If we can show that this is true to the series given to us in the question, then we have a bound on the absolute value of 𝑆 minus the 𝑛th partial sum. We’ll start by setting our sequence π‘Ž 𝑛 to be equal to three divided by six to the 𝑛th power plus 10.

The first thing we need to do is show that our sequence π‘Ž 𝑛 is positive. We see that π‘Ž 𝑛 is equal to three divided by six to the 𝑛th power plus 10. And we can see that both three and 10 are positive numbers and six to the 𝑛th power is positive for all values of 𝑛. So, for all values of 𝑛, π‘Ž 𝑛 is just the quotient of two positive numbers. So, π‘Ž 𝑛 is a positive sequence.

Next, we need to show that our sequence π‘Ž 𝑛 is decreasing. We might be tempted to do this by setting the function 𝑓 of π‘₯ equal to three divided by six to the power of π‘₯ plus 10 and then checking the slope of this function. However, there’s actually a simpler method. We’ll just compare π‘Ž 𝑛 and π‘Ž 𝑛 plus one directly. We can see that both π‘Ž 𝑛 and π‘Ž 𝑛 plus one share the same positive numerator of three.

If we were to compare the denominators of these two expressions, we see that six to the power of 𝑛 plus one is bigger than six to the 𝑛th power. So, the denominator of π‘Ž 𝑛 plus one is a bigger positive number than the denominator of π‘Ž 𝑛. And if we divide a positive number by a bigger positive number, that makes this number smaller. So, π‘Ž 𝑛 plus one is smaller than π‘Ž 𝑛. So, we’ve shown that our sequence π‘Ž 𝑛 is decreasing.

Next, we need to show the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 is equal to zero. So, we want to show the limit as 𝑛 approaches ∞ of three divided by six to the 𝑛th power plus 10 is equal to zero. There’s a few different ways of seeing this. For example, we could take the constant three outside of our limit. And again, since our denominator is always positive, we can remove the plus 10 to make our denominator smaller. By removing this plus 10, we’ve made our positive denominator smaller. So, we’ve made every term in this new sequence bigger. So, we have that the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 is less than or equal to three times the limit as 𝑛 approaches ∞ of one divided by six to the 𝑛th power.

And we can see that this limit approaches zero. The numerator remains constant. However, the denominator is unbounded and approaches ∞. So, the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 is less than or equal to zero. And remember, we already showed that our sequence π‘Ž 𝑛 is a positive sequence. So, if every term in the sequence is positive, we have zero is less than or equal to the limit as 𝑛 approaches ∞ of π‘Ž 𝑛. And this means we can conclude that the limit as 𝑛 approaches ∞ of π‘Ž 𝑛 is equal to zero.

So, we’ve shown that all of the prerequisites are true. This means we’re now allowed to use our bound. Remember, the question wants our bound to be at most 10 to the power of negative eight. We can find a sufficient value of 𝑛 by choosing 𝑛 large enough that π‘Ž 𝑛 plus one is less than or equal to 10 to the power of negative eight. Since, in this case, the absolute value of 𝑆 minus our 𝑛th partial sum is less than or equal to π‘Ž 𝑛 plus one, but we chose 𝑛 large enough that this is less than or equal to 10 to the power of negative eight.

So, let’s find this value of 𝑛. We want three divided by six to the power of 𝑛 plus one plus 10 to be less than or equal to 10 to the power of negative eight. Both sides of this inequality are positive. So, we can take the reciprocal of both sides of the inequality. Well, we must be careful to flip the direction of our inequality. This gives us six to the power of 𝑛 plus one plus 10 all divided by three is greater than or equal to 10 to the eighth power. We can then multiply this inequality through by three. Next, we’ll subtract 10 from both sides of this inequality.

So, we have six to the power of 𝑛 plus one is greater than or equal to three times 10 to the power of eight minus 10. We want to find the value of 𝑛. So, we’ll take the log base six of both sides of this inequality. By using our laws of logarithms, the log base six of six to the power of 𝑛 plus one is just equal to 𝑛 plus one. Finally, we subtract one from both sides of this inequality. And if we evaluate this expression, we see it’s approximately equal to 9.89. The value of 𝑛 must be an integer. So, saying 𝑛 is greater than or equal to 9.89 is the same as saying that 𝑛 is greater than or equal to 10.

So, we’ve shown that all values of 𝑛 greater than or equal to 10 are sufficient. What do we mean by this? Well, we’ve shown the absolute value of 𝑆 minus our 10th partial sum is less than or equal to π‘Ž 11, which, in turn, is less than or equal to 10 to the power of negative eight. But remember, the question wants us to find the lowest value of 𝑛 which has this property. Which, as we said earlier, means we don’t only want 𝑛 to have this property, we need that 𝑛 minus one does not have this property. In other words, we want to reduce the number of terms in our approximation until the absolute value of 𝑆 minus this partial sum is bigger than 10 to the power of negative eight.

So, let’s clear some space and have a look at the absolute value of 𝑆 minus the ninth partial sum. We immediately run into a problem. We don’t actually know the value of 𝑆, and our series is not in a form which we can easily calculate the value. So, how are we going to approximate the absolute value of 𝑆 minus the ninth partial sum? We already know we can approximate the value of 𝑆 by taking a partial sum. So, what if instead of calculating the absolute value of 𝑆 minus the ninth partial sum, we instead found an approximation for 𝑆, which is incredibly accurate, for example, what if we’ve made our approximation accurate to 10 to the power of negative 10.

We can do the exact same series of calculations on this inequality which we did before. We end up with 𝑛 greater than or equal to the log base six of three times 10 to the 10th power minus 10 minus one, which we can calculate to be approximately 12.46. And since 𝑛 is an integer, saying 𝑛 is greater than or equal to 12.46 is the same as saying 𝑛 is greater than or equal to 13. So, the absolute value of 𝑆 minus our 13th partial sum is less than or equal to 10 to the power of negative 10.

We can approximate 𝑆 by the 13th partial sum to an incredibly accurate degree. So, instead of looking at the absolute value of 𝑆 minus the ninth partial sum, we can instead look at the absolute value of the 13th partial sum minus the ninth partial sum. And this is easier to calculate since the 13th partial sum minus the ninth partial sum is actually just equal to the 10th term in our series plus the 11th term in our series plus the 12th term in our series plus the 13th term in our series. It’s just the sum of four terms of our series.

So, there’re a few different ways of calculating these values. We could use the 𝛴 notation for our partial sums to calculate these values. However, we’re just going to calculate the values of these series directly. Calculating the 13th partial sum of this series to 11 decimal places, we get negative 0.13359006038. And remember, the 13th partial sum was our approximation for 𝑆. It’s accurate to 10 to the power of negative 10. But we want our approximations to be accurate to 10 to the power of negative eight. So, we’ll be mainly focused on the eighth decimal place of this approximation.

Let’s start by looking at the decimal expansion of our 10th partial sum. Remember, we know that this already works. We can see that the first seven terms in our decimal expansion agree with 𝑆 13. And if we look at the values around our eighth decimal place, we can see the absolute value of our 13th partial sum minus our 10th partial sum is less than or equal to 10 to the power of negative eight. But we already knew this would be the case since we showed that when 𝑛 is greater than or equal to 10, these values of 𝑛 are sufficient.

So, let’s take a look at the decimal expansion of our ninth partial sum. When we calculate our ninth partial sum, we can see that only the first six terms of the decimal expansion agree with 𝑆 13. We can actually see that this approximation is off by about 10 to the power of negative seven. In other words, approximating 𝑆 by the ninth partial sum will be off by more than 10 to the power of negative eight. And remember, this must be true since the absolute value of 𝑆 minus the 13th partial sum is less than or equal to 10 to the power of negative 10. So, in actual fact, the value of 𝑛 is equal to 10 was the lowest value of 𝑛 which guaranteed that this was true.

In conclusion, we’ve shown the lowest value of 𝑛 which guarantees that the 𝑛th partial sum of the series 𝑆 is equal to the sum from 𝑛 equals one to ∞ of negative one to the 𝑛th power times three. Divided by six to the 𝑛th power plus 10 differs from the infinite sum by 10 to the power of negative eight at most is 𝑛 is equal to 10.

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