Solve in english

The general term of the series is $a_n = \frac{1}{n(n+1)}$. We can express this term as a difference of two fractions:

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

To find $A$ and $B$, we multiply both sides by $n(n+1)$:

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

Set $n=0$: $1 = A(0+1) + B(0) \implies 1 = A$.
Set $n=-1$: $1 = A(-1+1) + B(-1) \implies 1 = -B \implies B = -1$.
So, the general term can be written as:

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

The partial sum $S_N$ is the sum of the first $N$ terms:

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

Let's write out the first few terms and the last term:

$$S_N = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{N} - \frac{1}{N+1}\right)$$

Notice that most of the terms cancel each other out. This is a telescoping series:

$$S_N = 1 - \cancel{\frac{1}{2}} + \cancel{\frac{1}{2}} - \cancel{\frac{1}{3}} + \cancel{\frac{1}{3}} - \cancel{\frac{1}{4}} + \dots + \cancel{\frac{1}{N}} - \frac{1}{N+1}$$

The partial sum simplifies to:

$$S_N = 1 - \frac{1}{N+1}$$

To find the sum of the infinite series, we take the limit of the partial sum as $N$ approaches infinity:

$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left(1 - \frac{1}{N+1}\right)$$

As $N \to \infty$, the term $\frac{1}{N+1}$ approaches 0.

$$\lim_{N \to \infty} \left(1 - \frac{1}{N+1}\right) = 1 - 0 = 1$$

The sum of the series is:

$$1$$