Math Question: (1/1)+(1/2)+(1/3)+(1/4)...

General Questions
#1

I’ve been trying to develop a simple formula, but have hit a rather large problem that I cannot seem to solve. I’ll spare you the details of what I’m trying to do and give you the problem straight off. I apologize for my basic mathematical notation.

How do I work out the sum of n terms in the sequence (1/1)+(1/2)+(1/3)…? For example, how do I find the total of (1/1)+(1/2)+(1/3)+(1/4)+(1/5)+(1/6) without actually adding each term separately? This is, of course, the same as (1^-1)+(2^-1)+(3^-1)… I am aware of formulas for arithmetic and geometric progressions, but this appears to be neither – is it an exponential progression, power law or whatever else? Am I forgetting some fundamental rule?

Whatever steps I’ve taken towards solving this, I always arrive back at the same basic problem. Factorial numbers (x!) seem to turn up often, but that might just be my repetitive thinking. I’ve given up trying to work this out myself.

Expert and maybe not-so-expert mathematicians of the Straight Dope message board, please help me out!

#2

I don’t believe there is a “closed form” equation that does what you want. Ya gotta just add the terms.

Trinopus

#3

http://mathworld.wolfram.com/HarmonicNumber.html

#4

In case you don’t feel like looking all the way through Achernar’s link, the sum of the first n reciprocals is approximately within a constant factor of ln(n).

#5

You won’t get a general equation without a factorial. There is no other way to get a common denominator. Lo siento mucho…

#6

Thanks for the links and information, especially for being so quick to reply! I’ve managed to finally complete the ‘simple’ formula I set out to create.

#7

One thing you need to be clear on: saying that f is approximately within a constant multiple of g does not mean that g is a good estimate for f. 10000000000000000000000000000000000000ln(n) + 3 is approximately within a constant multiple of ln(n), but the two have very different values.

#8

ultrafilter: H[sub]n[/sub] is not within a constant factor, it is withing an additive constant of ln n.

As to questions of the OP’s type, in addition to looking it up, solving using geometric series, etc., if all else fails with a sum, approximate it (from below and above) by an integral. The integral of 1/n is especially easy.

#9

Hmmm…isn’t this another way of writing out pi? Sorry - I forget the exact different factorials involved.

#10

Nope. The series diverges.

#11

Do I remember wrong? I thought the summation of 1/n didn’t converge to a limit.

#12

Hmm…apparently I misunderstood the ‘~’ notation. Let me go back and review that–it may be more general than I thought it was.

#13

What about something like:

lim (x->&) of SUM (1/x)?

The sum can be expanded with a formula, I believe, then you just take the limit as x goes to infinity.

#14

I’m no math expert but it seems like the area under the curve of the function should do the trick. In this case, the integral from 1 to n of (1/x) dx, which comes out to the ln n - ln 1.

Oh wait, did you only want this for integers?

If so, might substituting the step function do the trick?

#15

It’s a divergent series, so that doesn’t help much.

#16

My last response was to Trigonal Planar.

#17

No it doesn’t, and here’s a neat way to show it:

1/3 + 1/4 > 2 X (1/4) > 1/2

1/5 + 1/6 + 1/7 + 1/8 > 4 X (1/8) > 1/2

1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16 > 8 X (1/16) > 1/2

Continue grouping and you show that the series is > 1/2 + 1/2 + 1/2 …

#18

Karl Gauss, I believe you, but I must ask: I ain’t a mathematician, but it sure looks like it should converge, or you should be able to take the limit. I mean, each additional term is adding just a smaller and smaller piece to the total. You’d think it’d reach a limit of some value as n grows larger and larger in the expression “1/n”.

I could express the OP’s question as a summation notation of 1/n as n ranges from 1 to infinity, couldn’t I? Then, couldn’t I find the limit as n -> infinity?

Please feel free to explain in greater detail where my thinking may be wrong. But, please be gentle! - Jinx

#19

I hear you, but it just isn’t so. Consider this series:

1.1 + 1.01 + 1.001 + 1.0001 + …

Each piece added is smaller and smaller, but it’s definitely going to go off to infinity, right?

#20

Archernar, yes, you are correct! Thanks for the explanation…
And, along these lines, if you are a (male) math teacher, then you know you’ll never kiss the girl. You’ll get closer and closer, but you’ll never get there! See? So, who says women need to play hard to get? :wink: - Jinx