Alternernating series

Infinite series: Convergence tests

Alternernating series

We have seen that the harmonic series $\sum\frac 1n$ diverges to infinity. Therefore $\lim_{n\to\infty} a_n = 0$ does not imply the series $\sum a_n$ converges. However, in the special case of so-called alternating series this is true:

If $(a_n)_{n=1}^\infty$ is a decreasing sequence of positive numbers with $\lim_{n\to\infty} a_n = 0$ then the series $\sum_{n=1}^\infty (-1)^n a_n$ converges.

Write $s_n = \sum_{k=1}^n a_k$ for the partial sums. The subsequence $(s_{2n})_{n=1}^\infty$ is decreasing because $s_{2n} - s_{2n-2} = (-1)^{2n}a_{2n} + (-1)^{2n+1}a_{2n+1} = a_{2n} - a_{2n+1} \leq 0 \text.$ Furthermore, $s_{2n} \geq s_1$ so this subsequence is bounded from below and this implies it converges. The subsequence $(s_{2n+1})_{n=0}^\infty$ on the other hand is decreasing, but also bounded from above. Therefore this subsequence converges as well. Since $\lim_{n\to\infty}s_{2n+1}-s_{2n}=0$ the limits are the same. We conclude that the partial sums converge indeed.

The series $\sum_{n=1}^\infty \frac{(-1)^n}{n}$ converges according to the theorem.

The alternating sequence $\sum_{n=1}^\infty (-1)^n a_n$ does not necessarily converge if $(a_n)_{n=1}^\infty$ is not decreasing or is not positive.

For $a_n = \begin{cases} \frac{1}{n} &\text{if }n\text{ is even} \\ \frac{1}{n^2}&\text{if }n\text{ is odd} \end{cases}$ it holds that $\sum_{n=1}^\infty (-1)^n a_n = \infty$ .