What is the Name of this Mathematical Series?

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liometopum
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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.
 
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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" . Umm, actually, I think "harmonic series" refers to the infinite series of which this is the first n+1 terms.
 
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micromass said:
I don't think so. Since the series diverges, I don't think many will find it interesting.

The harmonic series diverges. I think you hurt its feelings :(okay I'm just feeling silly...
 
ArcanaNoir said:
The harmonic series diverges.
Just barely, though...
 
pmsrw3 said:
Just barely, though...

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.
 
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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?
 
ArcanaNoir said:
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?

Check http://en.wikipedia.org/wiki/Small_set_(combinatorics) :smile:
 
micromass said:
What does barely mean?
It diverges very, very slowly!
 
pmsrw3 said:
It diverges very, very slowly!

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 :smile:
 
micromass said:
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.

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

[tex]\sum_{n=1}^\infty \frac{1}{n^{1+\epsilon}} < \infty[/tex]
for any [itex]\epsilon > 0[/itex]! ;) (the exclamation point denotes excitement, not a factorial! =P)
 
Maybe if you discover cool enough properties for the series, you can get to name it yourself. ;)
 
micromass said:
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 :smile:
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:

[tex] \begin{array}{l}<br /> \sum _{k=n}^{\infty } 1 \\<br /> \sum _{k=n}^{\infty } \frac{1}{k} \\<br /> \sum _{k=n}^{\infty } \frac{1}{k \log (k)} \\<br /> \sum _{k=n}^{\infty } \frac{1}{k \log (k) \log (\log (k))} \\<br /> \sum _{k=n}^{\infty } \frac{1}{k \log (k) \log (\log (k)) \log (\log (\log (k)))} \\<br /> ...<br /> \end{array}[/tex]

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.
 
Mute said:
If I increase the exponent on the series just "barely", though, it converges!

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

And yet

[tex]\sum{\frac{1}{n^{1+\frac{1}{n}}}}[/tex]

also diverges. So I can increase the exponent a bit, and it will still diverge!
 
pmsrw3 said:
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:

[tex] \begin{array}{l}<br /> \sum _{k=n}^{\infty } 1 \\<br /> \sum _{k=n}^{\infty } \frac{1}{k} \\<br /> \sum _{k=n}^{\infty } \frac{1}{k \log (k)} \\<br /> \sum _{k=n}^{\infty } \frac{1}{k \log (k) \log (\log (k))} \\<br /> \sum _{k=n}^{\infty } \frac{1}{k \log (k) \log (\log (k)) \log (\log (\log (k)))} \\<br /> ...<br /> \end{array}[/tex]

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.

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 :smile:
 
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micromass said:
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 :smile:
Really! :-)