# The harmonic sequence

• Nov 10th 2008, 06:36 AM
Showcase_22
The harmonic sequence
Quote:

Give, with reasons, a value of N for which $\displaystyle 1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{N} \geq 10$
I knew that the number of terms that are greater than $\displaystyle \frac{1}{2}$ each time follow the triangle numbers.

$\displaystyle 1+\frac{n}{2} \geq 10$

$\displaystyle \frac{n}{2} \geq 9$

$\displaystyle n \geq 18$ groups of $\displaystyle \frac{1}{2}$.

Therefore since the formula for the triangle number is $\displaystyle \frac{1}{2}n(n+1)$:

$\displaystyle \frac{1}{2}(18)(18+1)=171$

Hence N=171
• Nov 10th 2008, 10:53 AM
Laurent
Quote:

Originally Posted by Showcase_22
Hence N=171

I'm not sure this is right since someone else (who is better at maths than me) got a number in the 200s.

I don't know this result about triangular numbers. But first of all, I don't agree with your value of $\displaystyle N$.
Perhaps you've already learned that $\displaystyle \sum_{k=1}^N \frac{1}{k}\sim_N \ln N$ or, more precisely, $\displaystyle \left(\sum_{k=1}^N\frac{1}{k}\right)-\ln N \to_N\gamma$ where $\displaystyle \gamma\simeq 0.577$ is called the Euler constant. Due to that, the least possible number $\displaystyle N$ you are looking for is probably near $\displaystyle e^{10-0.577}\simeq 12370$.
In fact, I checked with Maple that the least $\displaystyle N$ is 12367, so this estimate is pretty precise...

There is a nice (classical) method on this wikipedia page that gives you a loose choice for $\displaystyle N$: $\displaystyle \sum_{k=1}^{2^n}\frac{1}{k}\geq 1+\frac{n}{2}$, hence $\displaystyle N=2^{18}=262144$ is OK. But not very close to the least possible choice...
• Nov 10th 2008, 12:45 PM
Showcase_22
sorry but what's "Maple"?

I'm going to try and find a way to get 12367 algebraically somehow. Is there a way?
• Nov 10th 2008, 12:51 PM
Mathstud28
Quote:

Originally Posted by Showcase_22
sorry but what's "Maple"?

I'm going to try and find a way to get 12367 algebraically somehow. Is there a way?

Maple is synonomous with Matchad. They are mathematics programs analgous to Mathematica. And for some reason I cannot see Lauren'ts post, but I assume he used the fact that for sufficiently large $\displaystyle N$: $\displaystyle \sum_{n=1}^{N}\frac{1}{n}\approx\ln\left(N\right)+ \gamma$. If that is what he did, that is what I would do, I am not quite sure about the triangle numbers thing. If he did not do it that way I would try it.

EDIT: FYI this should be labeled the "Harmonic SERIES". I only say because some instructors would mark off for the confusion.
• Nov 10th 2008, 01:24 PM
CaptainBlack
Quote:

Originally Posted by Mathstud28
Maple is synonomous with Matchad. They are mathematics programs analgous to Mathematica.

Matlab is a primarily numerical computation system
Mathematica and Maple are primarily symbolic packages

However you can get a symbolic package for Matlab which last time I looked used the Maple kernel, and Maple and mathematica do no numerics.

(at least two of these are Trojan horses, sold cheaply to students so they will demand them when they start work, but the price to comercial users are punative)

CB
• Nov 10th 2008, 01:26 PM
Mathstud28
Quote:

Originally Posted by CaptainBlack

Matlab is a primarily numerical computation system
Mathematica and Maple are primarily symbolic packages

However you can get a symbolic package for Matlab which last time I looked used the Maple kernel, and Maple and mathematica do no numerics.

CB

Oops, thank you. I did mean "Mathcad"
• Nov 11th 2008, 03:44 AM
Laurent
Quote:

Originally Posted by Showcase_22
sorry but what's "Maple"?

I'm going to try and find a way to get 12367 algebraically somehow. Is there a way?

I think it would be hard to find exaclty 12367 without proceding to the addition of all terms with a sufficient precision, which means doing it by hand, or using Matlab or whatever (the nice thing with Maple (even though it is indeed not designed for numerics) is that it can have an arbitrary precision; this certainly isn't necessary here, but for very large $\displaystyle N$, or slower diverging series, this could be important)

The asymptotic expansion I gave for $\displaystyle \sum_{k=1}^N\frac{1}{k}$ can not be used as such for a proof, because it is asymptotic, but we can use the idea that is involved when proving it: comparison with an integral.
You can write $\displaystyle \sum_{k=1}^N \frac{1}{k}\geq \sum_{k=1}^N \int_k^{k+1}\frac{dt}{t}=\int_1^{N+1}\frac{dt}{t}= \ln(N+1)$, hence this is greater than 10 as soon as $\displaystyle N\geq e^{10}-1$, which means $\displaystyle N\geq 22026$. Not extremely accurate though, but much better than my first upper bound. I don't think the teacher expects more from you.
• Nov 11th 2008, 04:47 AM
Showcase_22
I thought a while back that my teacher didn't expect more from me. I just thought this Euler constant looked pretty good. I also suppose that if you know you can make your answer more accurate (in this case a LOT more accurate) then you really should do it.

(btw, there is nothing connecting the triangle numbers with the harmonic series, I was just confusing the previous question with this one).

Thanks guys! I'll download a maths computing programme now so I have some evidence when I talk to my supervisor tomorrow.