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Understand the logic behind the subtraction of 0

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  • #1
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Homework Statement


Problem:
Prove that if ##lim_{n→+∞} s_n## = ∞ and ##s_n## ≠ 0 for all n, then ##lim_{n→+∞} 1/s_n## = 0.

Solution:
Consider any ϵ > 0. Since ##lim_{n→+∞} s_n## = ∞, there exists some positive integer m such that, whenever n ≥ m,

##|s_n|## > 1/ϵ and, therefore, ##|##1/s_n## - 0|## = ##|1/s_n|## < ϵ. So, ##lim_{n→+∞} 1/s_n## = 0.

(I also attached the problem and its solution in the TheProblemAndSolution.jpg file.)

Homework Equations


Limits.

The Attempt at a Solution


I understand the n ≥ m part and, I suppose the ϵ > 0 and ##|s_n|## > 1/ϵ parts wouldn't throw me off if I understood the rest of the proof, but, I don't understand the logic behind the subtraction of 0 within the absolute value and why that is even needed because, I could just multiply both sides by ϵ/|##s_n##| to get |1/##s_n##| < ϵ. Lastly, I don't understand how one can conclude that ##lim_{n→+∞} 1/s_n## = 0 by looking at |1/##s_n##| < ϵ. To me, all that says is that ##lim_{n→+∞} 1/s_n## is less than some positive number so, it can be a smaller positive number or a negative number … this interval includes 0 but there are many other values in the interval so, I cannot see how one could simply conclude that ##lim_{n→+∞} 1/s_n## is 0.

Any help in understanding what's going on with this problem would be really appreciated!
 

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  • #2
Office_Shredder
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The subtraction is necessary because to prove that the limit of 1/sn is equal to zero, you have to show that |1/sn - 0| is very small. You should write down the definition of a limit and see what it says with regards to proving that a limit is zero.
 
  • #3
Ray Vickson
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Homework Statement


Problem:
Prove that if ##lim_{n→+∞} s_n## = ∞ and ##s_n## ≠ 0 for all n, then ##lim_{n→+∞} 1/s_n## = 0.

Solution:
Consider any ϵ > 0. Since ##lim_{n→+∞} s_n## = ∞, there exists some positive integer m such that, whenever n ≥ m,

##|s_n|## > 1/ϵ and, therefore, ##|##1/s_n## - 0|## = ##|1/s_n|## < ϵ. So, ##lim_{n→+∞} 1/s_n## = 0.

(I also attached the problem and its solution in the TheProblemAndSolution.jpg file.)

Homework Equations


Limits.

The Attempt at a Solution


I understand the n ≥ m part and, I suppose the ϵ > 0 and ##|s_n|## > 1/ϵ parts wouldn't throw me off if I understood the rest of the proof, but, I don't understand the logic behind the subtraction of 0 within the absolute value and why that is even needed because, I could just multiply both sides by ϵ/|##s_n##| to get |1/##s_n##| < ϵ. Lastly, I don't understand how one can conclude that ##lim_{n→+∞} 1/s_n## = 0 by looking at |1/##s_n##| < ϵ. To me, all that says is that ##lim_{n→+∞} 1/s_n## is less than some positive number so, it can be a smaller positive number or a negative number … this interval includes 0 but there are many other values in the interval so, I cannot see how one could simply conclude that ##lim_{n→+∞} 1/s_n## is 0.

Any help in understanding what's going on with this problem would be really appreciated!
If ##s_n \to +\infty## then, certainly, ##s_n > 0## for all n greater than some value m, so ##1/s_n > 0## for all n > m. There is no need to look separately at ##1/|s_n|##.
 
  • #4
s3a
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Ray Vickson, just to be pedantic, you meant n ≥ m rather than n > m, right? Not that n > m is wrong but including n = m is perfectly fine, right?

As for the rest of what you guys said, is this ( https://en.wikipedia.org/wiki/(ε,_δ)-definition_of_limit#Worked_Example ) the same kind of problem as the one I am dealing with except mine is more abstract and is a function of n rather than x?

What the solution to the problem that I posted was doing (when it skipped a step) was 0 < | 1/##s_n## - limit_which_is_0 | = 1/##s_n## - limit_which_is_0 < some_FIXED_number_ϵ_which_bounds_the_function? Why would one need to subtract the limit at all? If the limit is fixed then, couldn't ϵ still be a fixed number larger than the (absolute value of the) function without the limit subtracted such that the function is still bounded?

To me, all that ϵ stuff would just be proving that the limit of 1/##s_n## converges to some fixed value but not that it converges to 0, specifically. Why/how is this proof satisfactory for saying that the function 1/##s_n## converges to a value of 0?
 
  • #5
Ray Vickson
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Ray Vickson, just to be pedantic, you meant n ≥ m rather than n > m, right? Not that n > m is wrong but including n = m is perfectly fine, right?

As for the rest of what you guys said, is this ( https://en.wikipedia.org/wiki/(ε,_δ)-definition_of_limit#Worked_Example ) the same kind of problem as the one I am dealing with except mine is more abstract and is a function of n rather than x?

What the solution to the problem that I posted was doing (when it skipped a step) was 0 < | 1/##s_n## - limit_which_is_0 | = 1/##s_n## - limit_which_is_0 < some_FIXED_number_ϵ_which_bounds_the_function? Why would one need to subtract the limit at all? If the limit is fixed then, couldn't ϵ still be a fixed number larger than the (absolute value of the) function without the limit subtracted such that the function is still bounded?

To me, all that ϵ stuff would just be proving that the limit of 1/##s_n## converges to some fixed value but not that it converges to 0, specifically. Why/how is this proof satisfactory for saying that the function 1/##s_n## converges to a value of 0?
Well, if n > m we certainly have n ≥ m+1, so how we write it is absolutely, 100% irrelevant.

I don't know why you are subtracting the limit (0 in this case), nor do I see why you want to use the absolute value in this case. You are just making things more cluttered than they need to be.

There seems to be something you are not "getting" here, and the only advice I will offer is that you sit down, relax and think things through thoroughly.
 
  • #6
s3a
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Well, if n > m we certainly have n ≥ m+1, so how we write it is absolutely, 100% irrelevant.

I don't know why you are subtracting the limit (0 in this case), nor do I see why you want to use the absolute value in this case. You are just making things more cluttered than they need to be.

There seems to be something you are not "getting" here, and the only advice I will offer is that you sit down, relax and think things through thoroughly.
Actually, that solution is not mine, it's my book's.

Also, the limit is 0 and the link to Wikipedia I gave says that you need to subtract the limit from the function and put all that in an absolute value, IN GENERAL, which, like you said, is not necessary in this case.

About the n and m situation, you confirmed what I was thinking. :)

I did think through this thoroughly and, I just see this as a way to confirm convergence to some unknown, fixed value but, I'm trying to figure out how this confirms convergence TO A SPECIFIC VALUE (which in this case is 0 - because the limit of 1/##s_n##, as n approaches infinity, is 0.
 
  • #7
Office_Shredder
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No, you do need to subtract the limit. For example if sn converged to 3, and you considered |1/sn|, you would never be able to show it's smaller than a small epsilon because it would be approximately 1/3! If |1/sn| is smaller than any arbitrary epsilon as n gets big, then it has to be converging to zero - if it was converging to L, then |1/sn| would be approximately equal to L, and in particular would NOT be smaller than any arbitrary epsilon. In that case |1/sn-L| would be smaller than any arbitrary epsilon as n gets big.
 
  • #8
Ray Vickson
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No, you do need to subtract the limit. For example if sn converged to 3, and you considered |1/sn|, you would never be able to show it's smaller than a small epsilon because it would be approximately 1/3! If |1/sn| is smaller than any arbitrary epsilon as n gets big, then it has to be converging to zero - if it was converging to L, then |1/sn| would be approximately equal to L, and in particular would NOT be smaller than any arbitrary epsilon. In that case |1/sn-L| would be smaller than any arbitrary epsilon as n gets big.
Convergence to non-zero is just a special case of convergence to 0; after all, if ##a_n \to L## then ##a_n - L \to 0.## A non-negative sequence ##s_n## converges to 0 if, for any given ##\epsilon > 0## there is an ##N## such that ##n \geq N \Rightarrow s_n < \epsilon.## Of course, if the sequence ##s_n## happened to not be ≥ 0 for all n, then we would need to use the absolute value instead. Here, the needed value of ##N## depends on ##\epsilon,## so should perhaps be written as ##N(\epsilon),## but this can be taken as implied, since we choose ##\epsilon## first, then find ##N##. This question was asking about convergence to 0, so subtracting the limit is NOT necessary: when you subtract 0 you do not change anything.
 
  • #9
s3a
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Ray Vickson, so, basically, you're saying that it's always unnecessary to subtract the limit (even when it's non-zero) because it doesn't matter to what value ##s_n## converges to as long as it's to a value larger than 1/ϵ, but infinitely close to it, right?

So, why do various sources, such as Wikipedia, subtract the limit?

As for the absolute value, I could totally see how one would put it in the beginning and then just remove it as a statement to the positive nature of the sequence. I don't think I'm having any trouble at all grasping the absolute value thing.

Also, I think I now see why the limit at the end of the proof is equal to zero. Is it because in the inequality, n ≥ m, n is any arbitrary large finite number (that is at least as large as m – if n is at least as large as m, then 1/##s_n## < ϵ) such that when you put that limit in front of it n is no longer an arbitrary large finite number but is now an infinite number?

Lastly, is the reason why the author started with |##s_n##| > 1/ϵ just so that it gets conveniently rearranged so that ϵ is on the numerator when we want to analyze |1/##s_n##| such that having started with |##s_n##| > ϵ instead would have also been fine? (If you want, you can ignore the absolute value signs. :) )
 

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