Telescoping sum of series

Jan 2010
142
0
Consider \(\displaystyle a_n=\frac{1}{n^2+n}\)

Find \(\displaystyle S_3, S_6, and S_10 \)

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

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

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

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

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

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


is that right?
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
Consider \(\displaystyle a_n=\frac{1}{n^2+n}\)

Find \(\displaystyle S_3, S_6, and S_10 \)

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

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

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

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

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

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


is that right?
First, use Partial Fractions to simplify \(\displaystyle \frac{1}{n^2 + n} = \frac{1}{n(n + 1)}\).


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

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


Then \(\displaystyle A(n + 1) + Bn = 1\)

\(\displaystyle An + A + Bn = 0n + 1\)

\(\displaystyle (A + B)n + A = 0n + 1\).


So \(\displaystyle A = 1\) and \(\displaystyle A + B = 0\).

So \(\displaystyle B = -1\).


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



So for your series:

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


Notice that if we expand this out...

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

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

So the sum is \(\displaystyle 1 - \frac{1}{n + 1}\).



What do you think \(\displaystyle S_3, S_6\) and \(\displaystyle S_{10}\) equal?


P.S. A telescopic series is a series where everything except the first and last terms cancel.
 
Jan 2010
142
0
First, use Partial Fractions to simplify \(\displaystyle \frac{1}{n^2 + n} = \frac{1}{n(n + 1)}\).


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

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


Then \(\displaystyle A(n + 1) + Bn = 1\)

\(\displaystyle An + A + Bn = 0n + 1\)

\(\displaystyle (A + B)n + A = 0n + 1\).


So \(\displaystyle A = 1\) and \(\displaystyle A + B = 0\).

So \(\displaystyle B = -1\).


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



So for your series:

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


Notice that if we expand this out...

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

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

So the sum is \(\displaystyle 1 - \frac{1}{n + 1}\).



What do you think \(\displaystyle S_3, S_6\) and \(\displaystyle S_{10}\) equal?


P.S. A telescopic series is a series where everything except the first and last terms cancel.
\(\displaystyle S_3=\frac{3}{4}, S_6=\frac{6}{7} \) and \(\displaystyle S_10=\frac{10}{11} \)

right?