Infinitely many infinitely small numbers.

[JonF]

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

Equal: one, zero, or something else?

[master_coda]

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

[arildno]

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

IMHO..

[master_coda]

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

IMHO..

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

[JonF]

then does

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

[hello3719]

in the way it is written yes

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

[master_coda]

then does

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

Yes, because:

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

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

[Gokul43201]

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.

[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.

[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

[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.

[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.

[Hurkyl]

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.

[Integral]

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.

[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.

[master_coda]

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 \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.

[Gokul43201]

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

[JonF]

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?
\lim_{n \rightarrow \infty}\sum^{n}_{i=1} \frac{1}{n-1} = 1

Gokul43201 im not sure what you are asking…

[jcsd]

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…

Yes 1 is the limit here also.

[Gokul43201]

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

[arildno]

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


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

The limit of this expression is 1 when n goes to infinity.

[JonF]

Ok, how about this one. Since \sum_{i=1}^{2n}\frac{1}{n}=2 what does \lim_{n\rightarrow\infty} \sum_{i=1}^{2n} \frac{1}{n}=

[arildno]

Ok, how about this one. Since \sum_{i=1}^{2n}\frac{1}{n}=2 what does \lim_{n\rightarrow\infty} \sum_{i=1}^{2n} \frac{1}{n}=
\lim_{n\rightarrow\infty} \sum_{i=1}^{2n} \frac{1}{n}=2

[hello3719]

also 2 since lim n->infinity of 2 is 2 since it is a constant

[wisky40]

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?

[wisky40]

I'm wonder about your next question.

[JonF]

Sorry guys if this is starting to get a little redundant. But what would this equal?
\lim_{n\rightarrow\infty} \sum_{i=1}^{n^n} \frac{1}{n}

and would the previous series be any different if instead of "n" as the upper hand limit, it was just straight infinity?

[jcsd]

Well that's equal to n^{n-1} so it's obviously divergent.

[Gokul43201]

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.

[matt grime]

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?

\sum_{i=1}^N f = Nf

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.

[zetafunction]

HARMONIC series is divergent, but REGULARIZABLE

\sum_{n=1}^{\infty} \frac{1}{n} = - \frac{\Gamma '(1)}{\Gamma(1)}

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

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

here the Zeta fucntion is the Hurwitz one , the above is the definition of zeta-regularized determinat

[HallsofIvy]

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
Then don't worry about it
\frac{1}{\infty}= 0
is not true in standard analysis.

[arildno]

Then don't worry about it
\frac{1}{\infty}= 0
is not true in standard analysis.
This thread is 6 years old, Halls!