# Infinitely many infinitely small numbers.

1. May 26, 2004

### JonF

Does
$$\lim_{n \rightarrow \infty} \sum^{n}_{i=1} 1/n$$

Equal: one, zero, or something else?

2. May 26, 2004

### master_coda

The series is divergent. It doesn't equal anything.

3. May 26, 2004

### arildno

Really?
$$\sum_{i=1}^{n}\frac{1}{n}=1$$

IMHO..

4. May 26, 2004

### master_coda

Oh, good point. I missed that. Yes, the answer is 1.

5. May 26, 2004

### JonF

then does

$$\lim_{n \rightarrow \infty} \sum^{n}_{i=1} \frac{2}{n} = 2$$

6. May 26, 2004

### hello3719

in the way it is written yes

didn't you mean to put 2/i instead of 2/n ?

7. May 26, 2004

### master_coda

Yes, because:

$$\sum^{n}_{i=1} \frac{2}{n} = 2$$

This is just taking the limit of a constant as $n\rightarrow\infty$.

8. May 27, 2004

### Gokul43201

Staff Emeritus
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. May 27, 2004

### wisky40

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. May 27, 2004

### JonF

No typo.

These questions stem from a thread that was around a several weeks ago, where it was stated that $$\frac{1}{\infty}=0$$. My real question that I’ve been building up to is: how can $$\lim_{n \rightarrow \infty}\sum^{n}_{i=1} 0 = 1$$

11. May 27, 2004

### arildno

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. May 27, 2004

### matt grime

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. May 27, 2004

### Hurkyl

Staff Emeritus
And another thing in particular,

$$\lim_{n\rightarrow\infty} \sum_{i=1}^n 1/n$$

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.

Last edited: May 27, 2004
14. May 27, 2004

### Integral

Staff Emeritus
Jon,
Perhaps you are not aware of the errors in your notation.

What you wrote

$$\sum^n_{i=1} \frac 1 n$$

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
$$\sum^n_{i=1} \frac 1 i$$

This represents a finite sum to some unspecified upper limit, As long as n is finite the sum is finite. If n -> $$\infty$$ then the sum diverges.

Last edited: May 27, 2004
15. May 27, 2004

### matt grime

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. May 27, 2004

### master_coda

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 $\lim_{k\rightarrow\infty}(1+1/k)^k$. 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. May 27, 2004

### Gokul43201

Staff Emeritus
So, you're (JonF) asking "what is limiting value of 1 when n-->infinity" ???

18. May 27, 2004

### JonF

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

Gokul43201 im not sure what you are asking…

19. May 27, 2004

### jcsd

Yes 1 is the limit here also.

20. May 27, 2004

### Gokul43201

Staff Emeritus
when you sum a function that is independent of the summing variable, are you not just multiplying the function by 'n' ?