# Telescoping sum of series

• May 9th 2010, 12:33 AM
Anemori
Telescoping sum of series
Consider $a_n=\frac{1}{n^2+n}$

Find $S_3, S_6, and S_10$

$\sum,(\frac{1}{n}+\frac{1}{(n+1)})$
$n=\infty$

n=1 $S_1 = (\frac{1}{1}+\frac{1}{1+1}) = (1+\frac{1}{2})$

n=2 $S_2 = (\frac{1}{2}+\frac{1}{2+1}) = (\frac{1}{2}+\frac{1}{3})$

n=3 $S_3 = (\frac{1}{3}+\frac{1}{3+1}) = (\frac{1}{3}+\frac{1}{4})$

n=6 $S_6 = (\frac{1}{6}+\frac{1}{6+1}) = (\frac{1}{6}+\frac{1}{7})$

n=10 $S_10 = (\frac{1}{10}+\frac{1}{10+1}) = (\frac{1}{10}+\frac{1}{11})$

is that right?
• May 9th 2010, 01:11 AM
Prove It
Quote:

Originally Posted by Anemori
Consider $a_n=\frac{1}{n^2+n}$

Find $S_3, S_6, and S_10$

$\sum,(\frac{1}{n}+\frac{1}{(n+1)})$
$n=\infty$

n=1 $S_1 = (\frac{1}{1}+\frac{1}{1+1}) = (1+\frac{1}{2})$

n=2 $S_2 = (\frac{1}{2}+\frac{1}{2+1}) = (\frac{1}{2}+\frac{1}{3})$

n=3 $S_3 = (\frac{1}{3}+\frac{1}{3+1}) = (\frac{1}{3}+\frac{1}{4})$

n=6 $S_6 = (\frac{1}{6}+\frac{1}{6+1}) = (\frac{1}{6}+\frac{1}{7})$

n=10 $S_10 = (\frac{1}{10}+\frac{1}{10+1}) = (\frac{1}{10}+\frac{1}{11})$

is that right?

First, use Partial Fractions to simplify $\frac{1}{n^2 + n} = \frac{1}{n(n + 1)}$.

If $\frac{1}{n(n + 1)} = \frac{A}{n} + \frac{B}{n + 1}$

$\frac{1}{n(n + 1)} = \frac{A(n + 1) + Bn}{n(n + 1)}$.

Then $A(n + 1) + Bn = 1$

$An + A + Bn = 0n + 1$

$(A + B)n + A = 0n + 1$.

So $A = 1$ and $A + B = 0$.

So $B = -1$.

Therefore $\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}$.

$\sum_{i = 1}^n{\frac{1}{i(i + 1)}} = \sum_{i = 1}^n{\left(\frac{1}{i} - \frac{1}{i + 1}\right)}$.

Notice that if we expand this out...

$\sum_{i = 1}^n{\left(\frac{1}{i} - \frac{1}{i + 1}\right)} = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n + 1}\right)$

Can you see that every term except for the first and last cancels?

So the sum is $1 - \frac{1}{n + 1}$.

What do you think $S_3, S_6$ and $S_{10}$ equal?

P.S. A telescopic series is a series where everything except the first and last terms cancel.
• May 9th 2010, 11:31 AM
Anemori
Quote:

Originally Posted by Prove It
First, use Partial Fractions to simplify $\frac{1}{n^2 + n} = \frac{1}{n(n + 1)}$.

If $\frac{1}{n(n + 1)} = \frac{A}{n} + \frac{B}{n + 1}$

$\frac{1}{n(n + 1)} = \frac{A(n + 1) + Bn}{n(n + 1)}$.

Then $A(n + 1) + Bn = 1$

$An + A + Bn = 0n + 1$

$(A + B)n + A = 0n + 1$.

So $A = 1$ and $A + B = 0$.

So $B = -1$.

Therefore $\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}$.

$\sum_{i = 1}^n{\frac{1}{i(i + 1)}} = \sum_{i = 1}^n{\left(\frac{1}{i} - \frac{1}{i + 1}\right)}$.

Notice that if we expand this out...

$\sum_{i = 1}^n{\left(\frac{1}{i} - \frac{1}{i + 1}\right)} = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n + 1}\right)$

Can you see that every term except for the first and last cancels?

So the sum is $1 - \frac{1}{n + 1}$.

What do you think $S_3, S_6$ and $S_{10}$ equal?

P.S. A telescopic series is a series where everything except the first and last terms cancel.

$S_3=\frac{3}{4}, S_6=\frac{6}{7}$ and $S_10=\frac{10}{11}$

right?
• May 9th 2010, 03:38 PM
Prove It
Quote:

Originally Posted by Anemori
$S_3=\frac{3}{4}, S_6=\frac{6}{7}$ and $S_10=\frac{10}{11}$

right?

Correct.