# Name of a mathematical series

1. Jul 25, 2011

### liometopum

Is there a name for this series:

1/2 + 2/3 + 3/4 + 4/5 + 5/6 + 6/7 + 7/8 + 8/9 +....+ n/(n+1)

Thanks.

2. Jul 25, 2011

### micromass

Staff Emeritus
I don't think so. Since the series diverges, I don't think many will find it interesting.

3. Jul 25, 2011

### pmsrw3

It's equal to n + 1 - (1 + 1/2 + 1/3 + 1/4 + ... + 1/n+1). The thing in parens is the http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29" [Broken]. Umm, actually, I think "harmonic series" refers to the infinite series of which this is the first n+1 terms.

Last edited by a moderator: May 5, 2017
4. Jul 25, 2011

### ArcanaNoir

The harmonic series diverges. I think you hurt its feelings :(

okay I'm just feeling silly.....

5. Jul 25, 2011

### pmsrw3

Just barely, though...

6. Jul 25, 2011

### micromass

Staff Emeritus
What does barely mean? I can deleted infinitely many terms from the harmonic series and it will still diverge... I can make the terms much smaller and it will still diverge.

Last edited: Jul 25, 2011
7. Jul 25, 2011

### ArcanaNoir

You can divide it in half and it still diverges. I know that's not as cool as taking infinitely many terms from it, but then again, it kind of is the same thing...

My prof started telling me about how you can take the terms with "9, 99, 999" or maybe with that as an exponent, or something?... and make it converge. He didn't really lay it out though, just kind of said something in passing. Was this a baseless rumor or is there something like that?

8. Jul 25, 2011

### micromass

Staff Emeritus
Check http://en.wikipedia.org/wiki/Small_set_(combinatorics)

9. Jul 25, 2011

### ArcanaNoir

Spiffy :)

10. Jul 25, 2011

### pmsrw3

It diverges very, very slowly!

11. Jul 25, 2011

### micromass

Staff Emeritus
It divergence is in the order of log(n). While this is extremely slow for all applications, I can still easily find sequences that diverge 100000 times slower. I just want to make clear that "slow" is relative

12. Jul 25, 2011

### Mute

If I increase the exponent on the series just "barely", though, it converges!

$$\sum_{n=1}^\infty \frac{1}{n^{1+\epsilon}} < \infty$$
for any $\epsilon > 0$! ;) (the exclamation point denotes excitement, not a factorial! =P)

13. Jul 25, 2011

### Anonymous217

Maybe if you discover cool enough properties for the series, you can get to name it yourself. ;)

14. Jul 26, 2011

### pmsrw3

Sure. In fact, I could find a series that diverges infinitely more slowly, and then I could find another that diverges infinitely more slowly than that, and so, on, ad infinitum:

$$\begin{array}{l} \sum _{k=n}^{\infty } 1 \\ \sum _{k=n}^{\infty } \frac{1}{k} \\ \sum _{k=n}^{\infty } \frac{1}{k \log (k)} \\ \sum _{k=n}^{\infty } \frac{1}{k \log (k) \log (\log (k))} \\ \sum _{k=n}^{\infty } \frac{1}{k \log (k) \log (\log (k)) \log (\log (\log (k)))} \\ ... \end{array}$$

But those would be contrived series, made up just for the purpose of diverging slowly. The harmonic series is about as slowly diverging a series as you're likely to bump into, unless you go hunting for slowly diverging series.

I also had in mind the point Mute made: considering just series with terms of the form ip, p=-1 is the edge case.

15. Jul 26, 2011

### micromass

Staff Emeritus
And yet

$$\sum{\frac{1}{n^{1+\frac{1}{n}}}}$$

also diverges. So I can increase the exponent a bit, and it will still diverge!

16. Jul 26, 2011

### micromass

Staff Emeritus
Those series are not contrived. I've seen them popping up in probability theory. Fine, they're useless, but they do pop up from time to time

Last edited: Jul 26, 2011
17. Jul 26, 2011

Really! :-)