Infinitely many infinitely small numbers.

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In summary, the conversation involves a discussion on the summation of a series with a variable limit and the concept of taking limits inside and outside of summations. It is concluded that taking the limit of a value inside a summation and then applying the summation does not always result in the correct answer. The conversation also delves into the topic of infinitesimals and the correct notation for infinite sums. Ultimately, the group agrees that the limit of the summation \sum^{n}_{i=1} \frac{1}{n} is 1.
  • #1
JonF
621
1
Does
[tex]\lim_{n \rightarrow \infty} \sum^{n}_{i=1} 1/n [/tex]

Equal: one, zero, or something else?
 
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  • #2
The series is divergent. It doesn't equal anything.
 
  • #3
Really?
[tex]\sum_{i=1}^{n}\frac{1}{n}=1[/tex]

IMHO..
 
  • #4
arildno said:
Really?
[tex]\sum_{i=1}^{n}\frac{1}{n}=1[/tex]

IMHO..

Oh, good point. I missed that. Yes, the answer is 1.
 
  • #5
then does

[tex]\lim_{n \rightarrow \infty} \sum^{n}_{i=1} \frac{2}{n} = 2[/tex]
 
  • #6
in the way it is written yes

didn't you mean to put 2/i instead of 2/n ?
 
  • #7
JonF said:
then does

[tex]\lim_{n \rightarrow \infty} \sum^{n}_{i=1} \frac{2}{n} = 2[/tex]

Yes, because:

[tex]\sum^{n}_{i=1} \frac{2}{n} = 2[/tex]

This is just taking the limit of a constant as [itex]n\rightarrow\infty[/itex].
 
  • #8
I think that's a typo, and the "real" series that JonF wants to describe IS the harmonic series, which does diverge, as coda mentioned.

And then again, perhaps not...didn't see that the follow up post was also from JonF.
 
  • #9
I agree with Gokul43201 that this series diverges.
let's write it like this:
S= 1+(1/2)+(1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+ - - - +1/16)+ - -
from here it's evident that (1/3+1/4)>(1/4+1/4)=(1/2), (1/5+1/6+1/7+1/8)>(1/8+1/8+1/8+1/8)=(1/2) and so on...now a new smaller series it's been created, from here S'(2)=1+(1/2)
S'(4)=1+(1/2)+(1/2)=1+2(1/2)
S'(8)=1+(1/2)+(1/2)+(1/2)=1+3(1/2)
- - - - - - - - - - - -- - - - - - -
- - - - - - - - - - - - - - - - --
S'(2^k)=1+k(1/2), since S'(2^k) diverges and S'(2^k) <S therefore S also diverges.
 
  • #10
No typo.

These questions stem from a thread that was around a several weeks ago, where it was stated that [tex]\frac{1}{\infty}=0[/tex]. My real question that I’ve been building up to is: how can [tex]\lim_{n \rightarrow \infty}\sum^{n}_{i=1} 0 = 1[/tex]
 
  • #11
Those sums are not at all the same!
You must relinquish the idea once and for all that infinity is a real number.
This means, in particular, that aritmetic operations performed on real numbers cannot naively be used when dealing with infinities.
 
  • #12
who says lim (sum 0)=1?

no one here ought to say that; you are taking the limit inside the sum, then the limit of the sum, when you're not allowed, to since there's an n in the summand and in the limit.
 
  • #13
And another thing in particular,

[tex]\lim_{n\rightarrow\infty} \sum_{i=1}^n 1/n[/tex]

is not a sum of infinitely many infinitely small numbers; it is the limiting value of a sequence of finite sums each equal to 1.



In standard analysis, there is no such thing as an infinitessimal or adding an infinite collection of numbers; these ideas are merely conceptual tools.
 
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  • #14
Jon,
Perhaps you are not aware of the errors in your notation.

What you wrote

[tex] \sum^n_{i=1} \frac 1 n [/tex]

Is not the correct notation for an infinite sum, the index, that is the variable referenced BELOW the sigma needs to appear in the expression following the sigma. What you have shown consists of a single term, which would be the upper limit. You should write
[tex] \sum^n_{i=1} \frac 1 i [/tex]

This represents a finite sum to some unspecified upper limit, As long as n is finite the sum is finite. If n -> [tex] \infty[/tex] then the sum diverges.
 
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  • #15
JonF specifically said that his notation was correct and not the harmonic series. the sequence of sums is:
1
1/2+1/2 = 1
1/3+1/3+1/3 = 1
1/4+1/4+1/4+1/4 = 1
etc

however the error is thinking that you may let n go to infinity in the denominator independently of the limit of the sum.

If the index letter of the sum does not appear in the summand, then it is correctly interpeted as saying that you are just adding the summand to itself n times where n is the upper limit.
 
  • #16
matt grime said:
however the error is thinking that you may let n go to infinity in the denominator independently of the limit of the sum.

This is the heart of the problem. When you take the limit of something, you have to consider that something as a whole. People sometimes make a similar mistake when they consider [itex]\lim_{k\rightarrow\infty}(1+1/k)^k[/itex]. You can't take the limit of the value inside the brackets first and then apply the exponent and conclude that the result is 1. Similarly you can't just take the limit of a value inside a summation and then apply the summation. Sometimes that will give you the correct answer, but that does not work in general.

When you consider the summation as a whole, you can see that it has a value of 1 regardless of the value of n. So the limit as n->infinity is 1.
 
  • #17
So, you're (JonF) asking "what is limiting value of 1 when n-->infinity" ?
 
  • #18
master_coda said:
This is the heart of the problem. When you take the limit of something, you have to consider that something as a whole. People sometimes make a similar mistake when they consider . You can't take the limit of the value inside the brackets first and then apply the exponent and conclude that the result is 1. Similarly you can't just take the limit of a value inside a summation and then apply the summation. Sometimes that will give you the correct answer, but that does not work in general.

Thank you. That makes a lot of sense. But then would?
[tex]\lim_{n \rightarrow \infty}\sum^{n}_{i=1} \frac{1}{n-1} = 1[/tex]

Gokul43201 I am not sure what you are asking…
 
  • #19
JonF said:
Thank you. That makes a lot of sense. But then would?
[tex]\lim_{n \rightarrow \infty}\sum^{n}_{i=1} \frac{1}{n-1} = 1[/tex]

Gokul43201 I am not sure what you are asking…

Yes 1 is the limit here also.
 
  • #20
when you sum a function that is independent of the summing variable, are you not just multiplying the function by 'n' ?
 
  • #21
JonF said:
Thank you. That makes a lot of sense. But then would?
[tex]\lim_{n \rightarrow \infty}\sum^{n}_{i=1} \frac{1}{n-1} = 1[/tex]

Yes, because:
[tex]\sum^{n}_{i=1} \frac{1}{n-1}=\frac{1}{n-1}\sum^{n}_{i=1}1=\frac{n}{n-1}[/tex]

The limit of this expression is 1 when n goes to infinity.
 
  • #22
Ok, how about this one. Since [tex]\sum_{i=1}^{2n}\frac{1}{n}=2[/tex] what does [tex]\lim_{n\rightarrow\infty} \sum_{i=1}^{2n} \frac{1}{n}=[/tex]
 
  • #23
JonF said:
Ok, how about this one. Since [tex]\sum_{i=1}^{2n}\frac{1}{n}=2[/tex] what does [tex]\lim_{n\rightarrow\infty} \sum_{i=1}^{2n} \frac{1}{n}=[/tex]
[tex]\lim_{n\rightarrow\infty} \sum_{i=1}^{2n} \frac{1}{n}=2[/tex]
 
  • #24
also 2 since lim n->infinity of 2 is 2 since it is a constant
 
  • #25
a little discussion about this matter for John F.

What if I want to write the sum of (1/n-1) as 1/(n-1)+1/(n-1)+ - - - -so on, and then use the lim1/(n-1)with n-> to infinity+lim1/(n-1)with n-> to infinity+ - - - - =0+0+0+ - - - - - = 0.Further, I can say that (infin.../infin...) is not define and it could be any number I want.On the other hand, if you say that lim of (n/n)=1 when n goesto infinity,it's because (n/n)=1 and then you think that you can use the definition of limite afterwards by writing lim(1) when n-> to infinity =1.In general, for a number A element of the [R] lim of A when n-> to infinity. Then the question is what do you is to show any number? is that make sense?
 
  • #26
I'm sorry about misspelling your name.

I'm wonder about your next question.
 
  • #27
Sorry guys if this is starting to get a little redundant. But what would this equal?
[tex]\lim_{n\rightarrow\infty} \sum_{i=1}^{n^n} \frac{1}{n}[/tex]

and would the previous series be any different if instead of "n" as the upper hand limit, it was just straight infinity?
 
  • #28
Well that's equal to [itex]n^{n-1}[/itex] so it's obviously divergent.
 
  • #29
As I said before, you can get rid of the entire summation hoopla if your function does not involve the summing parameter. All you do is multiply and then find the limit.
 
  • #30
JonF, gokul has said several times, as have I, and probably others, that when you're doing a sum and the inddex of the sum doesn't appear in the summand (there is no i dependence here) then you're adding up a constant a certain number of times.

shall we make it explicit?

[tex] \sum_{i=1}^N f = Nf[/tex]

if the f in the summand has nothing to do with an i. It doesn't matter if it has anything to do with the upper limit, or the lower limit, it is a constant. You THEN take the limit as N tends to infinity.

Next time try thinking about your sum, you'll start to see what's going on better.
 
  • #31
HARMONIC series is divergent, but REGULARIZABLE

[tex] \sum_{n=1}^{\infty} \frac{1}{n} = - \frac{\Gamma '(1)}{\Gamma(1)} [/tex]

the idea is that Harmonic series is the logarithmic derivative (a=1) of the infinite product

[tex] \prod (n+a) [/tex] which can be 'regularized' to give [tex] e^{ - \zeta ' _{H} (0,a) [/tex]

here the Zeta function is the Hurwitz one , the above is the definition of zeta-regularized determinat
 
  • #32
JonF said:
No typo.

These questions stem from a thread that was around a several weeks ago, where it was stated that [tex]\frac{1}{\infty}=0[/tex]. My real question that I’ve been building up to is: how can [tex]\lim_{n \rightarrow \infty}\sum^{n}_{i=1} 0 = 1[/tex]
Then don't worry about it
[tex]\frac{1}{\infty}= 0[/tex]
is not true in standard analysis.
 
  • #33
HallsofIvy said:
Then don't worry about it
[tex]\frac{1}{\infty}= 0[/tex]
is not true in standard analysis.
This thread is 6 years old, Halls!
 

1. What are infinitely small numbers?

Infinitely small numbers, also known as infinitesimals, are numbers that are smaller than any positive real number but are still greater than zero. They are used in calculus and other branches of mathematics to represent quantities that are approaching zero, but never actually reach it.

2. How are infinitely small numbers different from zero?

Infinitely small numbers are different from zero because they have a non-zero value, even though it is incredibly small. Zero is a well-defined number that represents the absence of a quantity, while infinitesimals represent quantities that are approaching zero but never actually reach it.

3. Are infinitely small numbers real numbers?

Yes, infinitely small numbers are real numbers. They are part of the real number system and can be represented on a number line just like any other real number. However, they are not considered to be "normal" real numbers because they are smaller than any positive real number.

4. Can infinitely small numbers be used in calculations?

Yes, infinitesimals can be used in calculations, particularly in calculus. They are used to represent quantities that are approaching zero, and can be manipulated using the same rules as other real numbers. However, they cannot be used in all mathematical operations and must be handled carefully to avoid mathematical errors.

5. Why are infinitely small numbers important in mathematics?

Infinitely small numbers are important in mathematics because they allow us to describe and analyze quantities that are approaching zero, but never actually reach it. This is particularly useful in calculus, where infinitesimals are used to define derivatives and integrals. They also have applications in other areas of mathematics, such as in the study of limits and continuity.

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