It is known that

¥ å r=1

1/r2=p2/6

Then, what about

n å r=1

1/r

Does it converge to a value when n®¥?
And how to prove that it has a limit or not?

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+¼

Divide by 2:

L/2=1/2+1/4+1/6+¼

Subtract these two equations:

L/2=1/1+1/3+1/5+¼

Therefore:

1/1+1/3+¼ = 1/2+1/4+¼

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

2n-1 å i=1

1/i ³ n

(*)

[To show this use the fact that if i is in [2^n-1,2ⁿ-1] then i ³ 2^n-1 so 1/i £ 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+¼+1/n then:

S(n)-lnn

converges to a real number as n®¥. This number is called Euler's constant. So S(n) diverges at about the same rate as lnn.

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

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

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