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## Integral Calculus

### Course: Integral Calculus>Unit 5

Lesson 10: Alternating series error bound

# Alternating series remainder

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.B (LO)
,
LIM‑7.B.1 (EK)
By computing only the first few terms of an alternating series, we can get a pretty good estimate for the infinite sum.  See why.

## Want to join the conversation?

• What is an "error", in this case?
• R is the error, and it is added to your partial sum. The total sum will be between the partial sum, and the partial_sum + error. The error makes it that the answer is not exact. But you can get a small enough enough error by calculating a partial sum of more terms. Because the error will always be less then the first term that didn't make the cut for the partial sum, the bigger the number of terms you choose for the calculation of the partial sum, the smaller the error.
• how do we know how many terms to take? For instance, in the above video Sal took first four terms of the series. The approximation (or the error) can be made "better" if more terms are included? Please do help.
• Similar with estimation with integrals, taking a higher number to test with will increase the accuracy of the answer, however, I think for this purpose Sal just chose a lower number so the math was less complex. I suppose it depends on what you're looking for. If you're in a test and crunched for time go with a smaller number to make it simpler. If you're doing something in physics and aerospace engineering maybe kick it up a few numbers. :P
• What does a negative remainder mean?
• it means that the estimate is greater than the true value.
• Is there any method by which we can find the exact value of infinite series ?
• Some series have what is called a "closed form", where we can express them as a finite number of functions nested together, like ∛(2-√(5/3)).

However, in the grand scheme of things, these series are quite rare. It's because of careful cherry-picking on the part of teachers and textbook authors that you usually see sums that can be written neatly like this. Proving that a series converges is usually easier than finding out what it converges to.

But for series that do have a closed form, we often have to play it by ear. Geometric series can be expressed as a/(1-r), which is proven on Khan Academy. There are telescoping series, which are a type of alternating series where almost every term is subtracted from itself, leaving one or two terms and a bunch of zeroes.

There are other techniques for computing series, many of which can be found in places like solutions to IMO problems.
• Alternatively, if we chose to estimate the alternating series by S5 + R5, we could make the case that R5 is negative by the same logic of pairing each remaining term where a5 is more negative than a6, etc. Can we not use the integral test in this case, and does this mean that we must be mindful of the index chosen for partial sum and remainder in estimating alternating series?
• Let me know if you ever figured out this answer bcuz I have to take a BC calc test on it in a few days lol
(1 vote)
• I am not sure why Sal didn't compare and contrast what he did in this video to what he did in the previous video. To see the difference, I computed the upper and lower bounds and came up with a completely different range. Sal's upper bound was way better than mine; his lower bound was way worse than mine.

My upper bound was 115/144+ 1/4 (not +1/25)
My lower bound was 115/144 + .1/5 (not + 0)

While the inequality holds, it appears that using the integration method, I ended up with a worse upper bound. So knowing the the |error| is <= the magnitude of the first non-included term, why would one use the integration method?
(1 vote)
• Over time "previous video" changes due to Sal's amazing productivity, but if you are talking about the two videos on using integrals to place bounds on an infinite sum, then I don't think that method applies when you have an alternating series as you do in this video. He specified in the first of the pair of videos that the series needed to be `continuous, positive and decreasing` to use integrals as a bounding method. You can see that if you try to graph this series---you cannot apply that integral method because your values are jumping alternately above and below the x-axis. Every other value is negative. So, instead, we can use the tricks he has shown us to bound this wildly leaping creature.
• okay so I jumped ahead like a loooot, but I do have a question:
how come
a - b + c = a - (b - c)
thanks!!
(1 vote)
• If we start with a-(b-c), then we can distribute the negative sign to the terms inside the parentheses. So, a-(b-c) = a+(-b+c) = a-b+c.
• If the series at started at n=0, would the value at n=0 be the first term or does the first term only exist at n=0?
(1 vote)
• IF the series started at n = 0, we would have a real problem: the first term would be `undefined`, and therefore the sum would be undefined. ¹⁄₀ is a pretty effective showstopper without even getting into the (-1) multiplier