Do you have any examples of a telescoping series that diverges
@JKMath2 күн бұрын
The majority of telescoping series are convergent due to the nature of the terms collapsing. However, a common example of a telescoping series that diverges would be a series with terms of 1, -1, 1, -1 and so on where the terms oscillate between -1 and 1. Since -1 + 1 = 0, you can see how the terms would collapse, however, as you look at more and more terms the partial sums oscillate between 0 and 1. 0 for when you have an even number of terms, and 1 when you have an odd number of terms. So because of this oscillation of the partial sums, the sum of the entire series does not converge to a single value. The series diverges despite the "telescoping" terms. Hope this helps!
@AlishaMayo-p4zАй бұрын
for the first example why did we go from a(sub n+1) then a (sub n). I thought it was the other way around?
@JKMathАй бұрын
I'm not sure I follow your question. Can you give me a timestamp for the part you are confused with? I'd be happy to help!
@adriennecui351Ай бұрын
I'm a bit confused as to why, for some S sub n values, you added a sub n, while for others, you added a sub n-4 or a sub n-1 before starting to cancel out the values.
@JKMathАй бұрын
How many terms I wrote down depended on how the terms collapse for the series. You don't have to write them all out like I did, I wrote them out to show visually how the terms collapse. What you want to do is look for a pattern in how they collapse with the first couple terms, and that will help you know how many you want to look at towards the end. So in example 1 of this video, the terms collapse immediately, there is a -1/4 and then right after that is 1/4 and then -1/5 and 1/5 and so on. There are no values in between, so clearly the result will just be the first value minus the second value of the final term. Same thing happens in example 2. But then in example 3 things are slightly different. There is a "delay" in the terms collapsing. there is -1/3, but then 1/2 and -1/4 before we see positive 1/3, so because of this delay we will need to look at more terms than in the other examples to see what will be left over. In this case, there will be 2 values left over, one from each of the first 2 terms, and one from each of the last 2 terms. Does that make sense? You just want to look for a pattern. Hope this helps!
@benhogervorst2 ай бұрын
No examples where n > 1?
@JKMath2 ай бұрын
The vast majority of series you work with will either begin at n=0 or n=1. Even if a telescoping series begins at a value of n greater than 1, the procedure remains the same. You can see visually by looking at the terms, that if you take out (for example) the n=1 and n=2 terms so it starts at n=3, that the rest of the terms will still have values that collapse/telescope. The value of n, or the term you start with has no effect on the process of determining the convergence of a telescoping series, however the sum may be slightly different, depending on how many terms you cut out.
@benhogervorst2 ай бұрын
@@JKMath I see that now, thanks for the clarification. Of course when you do homework problems, they always throw stuff in there that you realistically won't see in the real world which is why I was dealing with problems where n > 1