proof of alternating series test

The series has partial sum

S_{2n+2}=a_{1}-a_{2}+a_{3}-+...-a_{2n}+a_{2n+1}-a_{2n+2},

where the $a_{j}$ ’s are all nonnegative and nonincreasing. From above, we have the following:

	$\displaystyle S_{2n+1}$	$\displaystyle=S_{2n}+a_{2n+1};$
	$\displaystyle S_{2n+2}$	$\displaystyle=S_{2n}+(a_{2n+1}-a_{2n+2});$
	$\displaystyle S_{2n+3}$	$\displaystyle=S_{2n+1}-(a_{2n+2}-a_{2n+3})$
		$\displaystyle=S_{2n+2}+a_{2n+3}$

Since $a_{2n+1}\geq a_{2n+2}\geq a_{2n+3}$ , we have $S_{2n+1}\geq S_{2n+3}\geq S_{2n+2}\geq S_{2n}$ . Moreover,

S_{2n+2}=a_{1}-(a_{2}-a_{3})-(a_{4}-a_{5})-\cdots-(a_{2n}-a_{2n+1})-a_{2n+2}.

Because the $a_{j}$ ’s are nonincreasing, we have $S_{n}\geq 0$ for any $n$ . Also, $S_{2n+2}\leq S_{2n+1}\leq a_{1}$ . Thus, $a_{1}\geq S_{2n+1}\geq S_{2n+3}\geq S_{2n+2}\geq S_{2n}\geq 0$ . Hence, the even partial sums $S_{2n}$ and the odd partial sums $S_{2n+1}$ are bounded. Also, the even partial sums $S_{2n}$ ’s are monotonically nondecreasing, while the odd partial sums $S_{2n+1}$ ’s are monotonically nonincreasing. Thus, the even and odd series both converge.

We note that $S_{2n+1}-S_{2n}=a_{2n+1}$ . Therefore, the sums converge to the same limit if and only if $a_{n}\to 0$ as $n\to\infty$ . The theorem is then established.

Title	proof of alternating series test
Canonical name	ProofOfAlternatingSeriesTest
Date of creation	2014-07-22 16:20:39
Last modified on	2014-07-22 16:20:39
Owner	Wkbj79 (1863)
Last modified by	pahio (2872)
Numerical id	13
Author	Wkbj79 (2872)
Entry type	Proof
Classification	msc 40A05