mathematics help: i need help on a math problem, please help me. - Help.com



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i need help on a math problem, please help me.

it’s a calc bc problem.

can someone give me an example of a telescoping series that diverges.

thanks

This open post was written 1 year, 9 months ago | V/U/S: 190, 3, 3 | Edit Post | Leave a reply | Report Post


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Creal Default offline Verified User (3 years) Long Term User Shouts: 2 #
Houston, TX, US | 1 year, 9 months ago (2 hours, 18 minutes after post)

I’m not 100% sure, but how about:
(n * (-1)^n) / n

I believe this diverges, but I’m not totally sure if this is considered a telescope series. Usually the telescope series only tells you if the series is convergent or if it’s not (but that doesn’t mean it’s divergent necessarily). This series evaluates to -1 and 1 infinitely, and is not convergent.

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chigozieji offline Unverified User #
An Unknown Location | 1 year, 7 months ago (2 months after post)

I want the know how to solve telecoping series

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Creal Default offline Verified User (3 years) Long Term User Shouts: 2 #
Houston, TX, US | 1 year, 7 months ago (2 months after post)

Well, I’m going to give a brief explanation, sorry I don’t have a real example on me. If the series follows the telescoping series rule, it is convergent. If it doesn’t, then it can be either divergent or convergent.

Pretty much, you start plugging the first couple of values for the infinte sum. If it is a telescoping series, you should notice a pattern where things are canceling. After you prove that they’re canceling, work the last couple of values of the sum (like “infinity - 1″, infinity {it’s acutally better practice to write this as the limit goes to infinity}. After you proof the cancelation again, take the remaining terms and add them together. This will give you a value of what the series is convergent to.

Pseudo-Ex: (Where each parathesis is a term after plugging in a value into the function and t goes to infinity)
f(x) = (1/n) - (1/(n+1))
(1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5)….
(1/(t-2) - 1/(t-1)) + (1/(t-1) - (1/t)) + (1/t + 1/(t+1))

As you can see in the example, the begging terms cancel, the -1/2 is canceled by adding the positive 1/2 in the next term. So, everything in this example cancels, except for the 1 (first term) and the 1/(t+1) (last term). With t approaching infinity, 1/(t+1) = 0 so this series is convergent to 1.

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