# Series and Sequences..deriving the formula

• Oct 31st 2009, 03:01 PM
kashifzaidi
Series and Sequences..deriving the formula
_______________________n
Consider the summation SUM ( (1/i) - (1/(i+1)) )
____________________ i=1

Derive a formula for this sum in terms of n.
n
Hint: SUM ( A(i) - A(i+1) )
i=1
= (A(1)-A(2)) + (A(2)-A(3)) + ... + (A(n)-A(n+1)) (who cancels??)
• Oct 31st 2009, 03:06 PM
mr fantastic
Quote:

Originally Posted by kashifzaidi
_______________________n
Consider the summation SUM ( (1/i) - (1/(i+1)) )
____________________ i=1

Derive a formula for this sum in terms of n.
n
Hint: SUM ( A(i) - A(i+1) )
i=1
= (A(1)-A(2)) + (A(2)-A(3)) + ... + (A(n)-A(n+1)) (who cancels??)

Write out the first few terms:

$\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + ....$.

Now simplify. What do you notice? What conclusion do you draw?
• Nov 5th 2009, 05:29 AM
kashifzaidi
Attachment 13692
I done the following according to my notes. My book says find the common thing and divide it by 2 on both sides. I cannot find any common. I dont know how do it futher..
Plz help me
This is a page from my book.
Attachment 13693
• Nov 5th 2009, 06:40 AM
Plato
The answer is simply $\sum\limits_{i = 1}^n {\left( {\frac{1}
{i} - \frac{1}
{{i + 1}}} \right)} = 1 - \frac{1}
{{n + 1}}$

These are known as collapsing sums.
Here only the first and last terms remain after subtraction.