Editing Alternating series (section)

== Alternating series test ==
{{main|Alternating series test}}

The theorem known as the "Leibniz Test" or the [[alternating series test]] states that an alternating series will converge if the terms {{math|''a<sub>n</sub>''}} converge to 0 [[monotonic function|monotonically]].

Proof: Suppose the sequence <math>a_n</math> converges to zero and is monotone decreasing. If <math>m</math> is odd and <math>m<n</math>, we obtain the estimate <math>S_n - S_m \le a_{m}</math> via the following calculation:
<math display="block">\begin{align}
S_n - S_m & =
\sum_{k=0}^n(-1)^k\,a_k\,-\,\sum_{k=0}^m\,(-1)^k\,a_k\ = \sum_{k=m+1}^n\,(-1)^k\,a_k  \\
& =a_{m+1} - a_{m+2} + a_{m+3} - a_{m+4} + \cdots + a_n\\
& = a_{m+1}-(a_{m+2}-a_{m+3}) - (a_{m+4}-a_{m+5}) - \cdots - a_n \le a_{m+1} \le a_{m}.
\end{align}</math>

Since <math>a_n</math> is monotonically decreasing, the terms <math>-(a_m - a_{m+1})</math> are negative. Thus, we have the final inequality: <math>S_n - S_m \le a_m</math>. Similarly, it can be shown that <math>-a_m \le S_n - S_m </math>. Since <math>a_m</math> converges to <math>0</math>, the partial sums <math>S_m</math> form a [[Cauchy sequence]] (i.e., the series satisfies the [[Cauchy criterion]]) and therefore they converge. The argument for <math>m</math> even is similar.