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 \(\sum_{n=1}^\infty (-1)^n a_n = \infty\) .