# Telescoping sum of series

#### Anemori

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

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

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.

#### Anemori

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

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

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?

#### Prove It

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

right?
Correct.