Missing step?

[mendhak]

Take a look at this:

http://www.brpreiss.com/books/opus6/html/img92.gif

How did the change in the limits (1 to n-1 => 1 to n) over the summation change (1/(i+1)) to ((1/i)-1)?

In other words, how did they get from step 2 to 3?

[zaza]

Hi mendhak,

It is because the -1 in step 3 is not part of the sum. For clarity it should be written: -1 + sum(1->n)[1/i].

Ways to tell:

1) You can test this fairly easily by bunging some numbers in.

2) The first sum is clearly 1/2 + 1/3 + ... + 1/n. The second is [1 + 1/2 + ... + 1/n] -1

3) If you want to prove it mathematically, do a substitution in the second sum of i = z+1. You then have sum(0->n-1)[1/{z+1}] -1
Evaluate the first term separately and you get:
1 + sum(1->n-1)[1/{z+1}] - 1 = sum(1->n-1)[1/{z+1}], which is what you had in the first place.


zaza