Evaluate the partial sum S_n and the sum S = lim as n

Evaluate the partial sum $S_n$ and the sum $S = \lim_{n \to \infty} S_n$ for the series: $$ \sum_{n=1}^{\infty} \frac{1}{n(n+1)}. $$

Step 1 · Partial fraction decomposition

The general term $a_n = \dfrac{1}{n(n+1)}$ is decomposed as:

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

Multiplying by $n(n+1)$ gives $1 = A(n+1) + Bn$. Setting $n=0$ gives $A=1$; setting $n=-1$ gives $B=-1$. Thus:

$$ a_n = \frac{1}{n} - \frac{1}{n+1}. $$

Step 2 · Determine the partial sum $S_n$

This is a telescoping series. The $n$-th partial sum is:

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

Writing out the terms:

$$ S_n = \left(\frac11-\frac12\right) + \left(\frac12-\frac13\right) + \left(\frac13-\frac14\right) + \cdots + \left(\frac1{n-1}-\frac1n\right) + \left(\frac1n-\frac1{n+1}\right). $$

The interior terms cancel in pairs, the hallmark of a telescoping series, leaving only the first positive term and the last negative term:

$$ S_n = \frac11 - \frac{1}{n+1} = 1 - \frac{1}{n+1}. $$

Step 3 · Calculate the sum $S$

The sum $S$ is the limit of $S_n$ as $n \to \infty$:

$$ S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right). $$

Since $\lim_{n \to \infty} \dfrac{1}{n+1} = 0$:

$$ S = 1 - 0 = 1. $$

Summary Table

Additional Notes

• The telescoping structure is what makes this series solvable in closed form; without the partial fraction split into $\dfrac1n-\dfrac1{n+1}$, there would be no obvious cancellation pattern.
• The partial sum $S_n = 1 - \dfrac1{n+1}$ is always less than $1$ for finite $n$, and increases monotonically toward $1$ as $n$ grows — consistent with a series of strictly positive terms converging from below.
• This result generalizes: for $\sum_{n=1}^\infty \frac{1}{n(n+k)}$ with integer $k\ge1$, the same telescoping idea gives a finite closed-form sum, of which this problem is the $k=1$ case.