Summation Notation With Variable Endpoint

Question

I have this equation:

$\sum_{n=1}^{k+1} \cfrac{1}{n(n+1)}$

The above is equivalent to: $\cfrac{1}{n(n+1)} | n=(k+1)$ Why is this true? Is it the result of integration? If so can someone show the work for it?

Answer 1

S = $\Sigma_{n=1}^{k+1}\frac{1}{n\left(n+1\right)}$

$\text{Partial Fraction Decomposition:}$

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

Multiply both sides by denominator of left side: $1 = A\left(n+1\right) + B\left(n\right)$

Plug in convenient value of n to isolate A: $n=0 \rightarrow A = 1$

Plug in convenient value of n to isolate B: $n=-1 \rightarrow B = -1$

$S = \Sigma_{n=1}^{k+1}\left(\frac{1}{n} - \frac{1}{n+1}\right)$

$\text{Telescoping Series:}$

$S = \Sigma_{n=1}^{k+1}\left(\frac{1}{n} - \frac{1}{n+1}\right) = \left( 1-\frac{1}{2}\right) + \left( \frac{1}{2}-\frac{1}{3}\right) + \left( \frac{1}{3}-\frac{1}{4}\right) + ... \left( \frac{1}{k}-\frac{1}{k+1} \right) + \left( \frac{1}{k+1}-\frac{1}{k+1+1}\right)$

Note how the "inner" terms cancel, just leaving the first term and the final term.

$S = 1 - \frac{1}{k+2} = \frac{k+1}{k+2}$