Closed form for $1/(n (n + 1))$

Question

What would the closed form be for $$ g(n) = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + ... + \frac{1}{n \cdot (n+1)}\;\;? $$

An example would be $g(3) = \frac{1}{2}+\frac{1}{6}+\frac{1}{12} = 0.75$

Answer 1

Use the idea of a telescoping sum.

$\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$.

Now sum it up from $k=1$ to $n$, and you shall see certain cancellations happening.

Answer 2

Use the idea of partial fractions to split the fraction apart:

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

Now we find $A$ and $B$:

$$1 = A(k+1) + Bk$$

When $k = -1$:

$$1 = -B, B = -1$$

When $k = 0$

$$1 = A$$

Therefore, our fraction now becomes:

$$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$$

This helps because now we can cancel out the terms, which I'm sure you can see now.

Comment if you have questions.

EDIT

If you expand the summation to $n$ terms, we can see that:

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

So the terms cancel out, and we get that:

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

Taking the limit as $n$ approached infinity:

$$\lim_{n\to\infty}S(n) = 1$$

Therefore the sum converges to $1$/