Analysis of the Series $ \sum_{k=1}^{\infty} \frac{(-1)^k}{k(k+1)} $

We’re asked to determine the convergence behavior of the series:

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

The general term is given by:

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

This is an alternating series since the sign of each term changes with $k$, due to the factor $(-1)^k$.

The Leibniz criterion for alternating series states that a series of the form

$$ \sum_{k=1}^\infty (-1)^k b_k $$

with $b_k \ge 0$, converges if both of the following conditions are met:

1. $b_k$ is eventually decreasing, that is, $b_{k+1} \le b_k$ for all $k \ge k_0$,
2. $\lim_{k \to \infty} b_k = 0$.

In our case, the positive part of the term is $ b_k = \frac{1}{k(k+1)} $, and clearly:

$$ \lim_{k \to \infty} \frac{1}{k(k+1)} = 0 $$

To verify the monotonicity, let’s compare $ b_k $ and $ b_{k+1} $:

$$ b_k = \frac{1}{k(k+1)} \quad \text{and} \quad b_{k+1} = \frac{1}{(k+1)(k+2)} $$

We observe that:

$$ \frac{1}{k(k+1)} > \frac{1}{(k+1)(k+2)} $$

which holds for all $k \ge 1$, so $\{b_k\}$ is decreasing.

Since both conditions of the Leibniz test are satisfied, the series converges.

Does the series converge absolutely?

To determine whether the series is absolutely convergent, we examine the series of absolute values:

$$ \sum_{k=1}^{\infty} \left| \frac{(-1)^k}{k(k+1)} \right| = \sum_{k=1}^{\infty} \frac{1}{k(k+1)} $$

We can simplify this by using a partial fraction decomposition:

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

To determine $A$ and $B$, we write:

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

Matching numerators:

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

This identity must hold for all $k$, so we equate coefficients:

$$ \begin{cases} A + B = 0 \\ A = 1 \end{cases} \quad \Rightarrow \quad \begin{cases} A = 1 \\ B = -1 \end{cases} $$

Thus,

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

Substituting back into the series, we get:

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

This is a classic telescoping series, which simplifies as follows:

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

Since the sum of the absolute values converges, the original series is absolutely convergent.

And so on.