$\sum_{r = 1}^{\infty} 1 / r$

By Woon Khang Tang on Friday, October 05, 2001 - 09:34 pm:

It is known that $\sum_{r = 1}^{\infty} 1 / r^{2} = π^{2} / 6$

Then, what about

$\sum_{r = 1}^{n} 1 / r$

Does it converge to a value when $n \to \infty$ ?
And how to prove that it has a limit or not?

By Michael Doré on Friday, October 05, 2001 - 09:50 pm:

Hi, This series doesn't converge - it diverges to infinity quite slowly.

Let us suppose it does converge to $L$ :

$L = 1 / 1 + 1 / 2 + 1 / 3 + \dots$

Divide by 2:

$L / 2 = 1 / 2 + 1 / 4 + 1 / 6 + \dots$

Subtract these two equations:

$L / 2 = 1 / 1 + 1 / 3 + 1 / 5 + \dots$

Therefore:

$1 / 1 + 1 / 3 + \dots = 1 / 2 + 1 / 4 + \dots$

But $1 / 1 > 1 / 2$ , $1 / 3 > 1 / 4$ , $1 / 5 > 1 / 6$ , ldots so the left hand side is clearly bigger than the right hand side so this is a contradiction, showing that our assumption that the series converges to $L$ is false. Can you see why this contradiction doesn't occur if the series doesn't converge?

There is an alternative method. Try to show that

$\sum_{i = 1}^{2^{n} - 1} 1 / i \geq n$ (*)

[To show this use the fact that if $i$ is in $[2^{n - 1}, 2^{n} - 1]$ then $i \geq 2^{n - 1}$ so $1 / i \leq 1 / 2^{n - 1}$ .]

Can you see why (*) gives the result immediately?

By the way, it is possible to get an idea about how fast the sequence tends to infinity. If you let $S (n)$ be $1 / 1 + 1 / 2 + \dots + 1 / n$ then:

$S (n) - \ln n$

converges to a real number as $n \to \infty$ . This number is called Euler's constant. So $S (n)$ diverges at about the same rate as $\ln n$ .

By Dave Sheridan on Monday, October 08, 2001 - 04:11 pm:

Another nice way to look at it is as follows.

1/3 > 1/4
1/5 > 1/8
1/6 > 1/8
1/7 > 1/8
etc

so that

1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+...

>

1+1/2+1/4+1/4+1/8+1/8+1/8+1/8+...

(In case it isn't obvious, this would be followed by 8 lots of 1/16, then 16 lots of 1/32. The smaller sum can be rewritten as
(1)+(1/2)+(1/4+1/4)+(1/8+1/8+1/8+1/8)+...
=1+1/2+1/2+1/2+...

So each time we add 1/2. This will increase without limit, although very slowly.

-Dave

By Woon Khang Tang on Tuesday, October 09, 2001 - 03:45 pm:

Thanks Dave, the proof u gave is very easy to understand and that's brilliant! How did you thought of that? WOW...

By Dave Sheridan on Wednesday, October 10, 2001 - 03:41 pm:

I've got to admit, it wasn't me who came up with it... It's a standard argument used in Cambridge undergrad lectures (or at least Andrew Thompson used it back when I was in my first year...)

-Dave