Terms of the Series 1/n

Date: 05/03/2001 at 14:24:18
From: Tony Steer
Subject: Series 1/n

With the series:

     1/1 + 1/2 + 1/3 + ... + 1/n

and given an answer that the series must exceed, how do I calculate 
the value of n? I am really stuck on this one; any help would be 
really appreciated. 

Thanks,
Tony

Date: 05/03/2001 at 16:25:47
From: Doctor Paul
Subject: Re: Series 1/n

Here's the idea. It follows directly from an intuitive proof that this 
series diverges. Let's look at the proof and then answer your 
question.

Notice the following:

                             1/3 + 1/4 > 1/4 + 1/4 = 1/2

     1/5 + 1/6 + 1/7 + 1/8 > 1/8 + 1/8 + 1/8 + 1/8 = 1/2

       1/9 + 1/10 + ... + 1/16 > 1/16 + ... + 1/16 = 1/2

Since the series never stops, I can always find another "group" of 
numbers whose sum is greater than one half. Thus, the series diverges.

Now let's say you want the series to be greater than three and you 
want to know how far out you have to go. Well, rewrite three as:

     3 = 1 + 1/2 + 1/2 + 1/2 + 1/2

       < 1 + 1/2 + [1/3+1/4] + [1/5+1/6+1/7+1/8] + [1/9+1/10+...+1/16]

So summing 1/n from n = 1 to 16 gives a result that is guaranteed to 
be greater than 3. Of course, the series gets larger than three before 
n = 16, but I don't see an easy way to figure out which value of n 
puts the sum of the series above three without computing it by hand. 
n = 11 is the first value that actually puts the sum greater than 3, 
but I found this by direct computation. The idea is to try to avoid 
this computation. And we have, but we've given up a bit of accuracy in 
doing so.

So by the argument above, we know that the sum will be greater than 3 
when you sum from n = 1 to 16.

What if we wanted the sum greater than four?

You would have to add two more "groups" of values - each of whose sum 
is greater than one half. Start with:

     1/17 + 1/18 + ... + 1/32 > 1/32 + 1/32 + ... + 1/32 = 1/2
and
     1/33 + 1/34 + ... + 1/64 > 1/64 + 1/64 + ... + 1/64 = 1/2

So the sum from n = 1 to 64 is guaranteed to be greater than four. 
n = 31 is the first value that makes the sum actually greater than 
four, but again - I had to use a calculator to find n = 31. I didn't 
need a calculator to prove that n = 64 works.

What if we wanted the sum greater than five?

Well, add two more "groups" - each greater than one half:

     1/65 + ... + 1/128 > 1/2
and
     1/129 + ... + 1/256 > 1/2

So if you sum from n = 1 to 256, you are guaranteed to be greater than 
five. The sum from n = 1 to 256 is actually greater than six. But how 
would you know this without a computing machine or hours of 
computation? By a much easier argument, we've shown that the sum from 
n = 1 to 256 must be greater than five.

Let's summarize our results:

     want sum greater than 2 -> sum from n = 1 to   4 = 2^2
     want sum greater than 3 -> sum from n = 1 to  16 = 2^4
     want sum greater than 4 -> sum from n = 1 to  64 = 2^6
     want sum greater than 5 -> sum from n = 1 to 256 = 2^8

Do you see a pattern?

it seems that if we want to guarantee that the sum will be greater 
than some number K, we need the following:

     want sum greater than K -> sum from n = 1 to 2^(2K-2)

I think that answers your question. If you don't follow the argument, 
or if you have additional questions, please write back.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/