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 ProofOfAlternatingSeriesTest 2014-07-22 16:20:39 2014-07-22 16:20:39 Wkbj79 (1863) pahio (2872) 13 Wkbj79 (2872) Proof msc 40A05