First, let us consider the convergence of the following series as n \to \infty: S = \sum_{i=1}^n (-1)^{i+1}(1/i) = 1 - 1/2 + 1/3 - 1/4 + \dots + (-1)^{n+1}(1/n)
Since \lim_{n \to \infty} (-1)^{n+1}(1/n) = 0, if S diverges, it must be that S \to \infty.
Suppose, for the sake of argument, that this series diverges. Then:
(I) S = (1 - 1/2) + (1/3 - 1/4) + \dots + (1/(2n) - 1/(2n+1)) Since the terms within each pair of parentheses are positive, S should tend toward +\infty.
(II) S = 1 + (-1/2 + 1/3) + (-1/4 + 1/5) + \dots + (-1/(2n) + 1/(2n+1)) + 1/(2n+2) Since each term within the parentheses is negative, and \lim_{n \to \infty}(1 + 1/(2n+2)) = 1, S should tend toward -\infty.
But S cannot tend to both +\infty and -\infty simultaneously, so the original assumption is incorrect. S converges.
From this, we can derive a convergence test for series:
Define the series \sum_{i=1}^{n}(-1)^{i}a_i or \sum_{i=1}^n(-1)^{i+1}a_i as an alternating series.
If the series \sum_{i=1}^{\infty}(-1)^{i}a_i satisfies:
1. \lim_{n \to \infty} a_n = 0
2. a_i \ge a_{i+1}
Then the series converges! The proof process is the same as the example above.
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