# What these number call

1. Jul 8, 2003

what the name of the numbers 1+ (1/2) + (1/3) + (1/4)+...+...+.....(1/n)...

can something one tell me wheather it has a finit sum or not

2. Jul 8, 2003

### Integral

Staff Emeritus
That is the http://www.jimloy.com/algebra/hseries.htm [Broken] series and it does NOT have a finite sum.

Last edited by a moderator: May 1, 2017
3. Jul 11, 2003

### STAii

I understand the proof provided in the page, but what i don't understand is, logically, why isn't there a sum.
You know, when i was first told about the sum of an infinite geometrical series, it first looked impossible, then i was told "since the numbers are getting smaller and smaller, they add up to give a number (not infinity)".
And here the numbers are getting smaller and smaller, but still, they don't sum up to a number, why is this ?

4. Jul 11, 2003

### mathman

simple proof

1/3+1/4>1/2
1/5+1/6+1/7+1/8>1/2
1/9+1/10+...+1/16>1/2
1/17+...+1/32>1/2
Keep this up and you get the harmonic series > 1+1/2+1/2+....

5. Jul 12, 2003

### STAii

This is the same proof found in the Integral's link, so this does not answer my question.
My question may seem a little weird, but anyone that feels (s)he can help by even giving a hint would be great.
Thanks.

6. Jul 12, 2003

### ahrkron

Staff Emeritus
This is probably where your problem resides. The fact that the numbes get "smaller and smaller" is not enough to insure convergence, as you just witnessed. They need to get smaller "fast enough", so to speak.

7. Jul 12, 2003

Staff Emeritus
It isn't enough that the terms get smaller, they have to get smaller fast enough that the sequence of partial sums converges. The geometric series 1 + 1/2 + 1/4 + 1/8 +... does this.
1 + 1/2 = 3/2
1 + 1/2 + 1/4 = 7/4
1+ 1/2 + 1/4 + 1/8 = 15/8

The partial sums are always of the form 2*2^n-1/2^n which is always less than 2, so the partial sums are bounded above and increasing, so they converge.

The harmonic series as the repeated proofs already posted show, doesn't do this, and this, not just the terms getting smaller is the true criterion for series convergence.

The terms getting smaller is a necessary condition for convergence; the series won't converge unless the do get smaller. But it is not a sufficient condition. Just terms getting smaller isn't enough by itself. The harmonic series is proof enough of that.

8. Jul 12, 2003

### STAii

How exactly does it mean 'fast enough' ?
Is there somekind of relation that must be between each number and the number after it so that it has sum (i am not only talking about geometrical series).
Thanks.

Last edited: Jul 12, 2003
9. Jul 12, 2003

### Lonewolf

A series is a sequence {an} with an assosciated partial sum {sn} where sn=&Sigma;i=1nai. For the series to be convergent, the sequence of partial sums must be convergent.

10. Jul 12, 2003

### Hurkyl

Staff Emeritus
You could use the Cauchy criterion:

The sum &Sigma;a(i) converges if and only iff:

limm,n&rarr;&infin; &Sigma;i=m..n a(i) = 0

This is equivalent to Lonewolf's definition for real numbers. (the Cauchy criterion fails in incomplete metric spaces)

11. Jul 13, 2003

### HallsofIvy

Staff Emeritus
One idea of "how fast" numbers in an infinite series must get smaller is the "ratio test":

The series [SIGMA] an converges if

lim |an|/|an+1 is less than 1

12. Jul 13, 2003

### ahrkron

Staff Emeritus
Just to clarify, I thik this is a sufficient condition, not a necessary one (1/n and 1/n^2 both fail the criterion, yet the latter is convergent).