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Let series $\displaystyle \sum _{n=n_0}^{+\infty}(-1)^na_n$ satisfy Leibniz convergence criterion. Let $\displaystyle r_n=\sum_{k=n+1}^{+\infty}(-1)^ka_k$. Prove that $r_n$ has the same sign as its first term and $|r_n|<a_{n+1}$.

Well, so far I used this fact several times but never consider proving it. In my opinion proofs of simple facts are not so obvious. I don't know how to precise show it.

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this should give you some ideas. –  David Mitra Apr 26 '12 at 15:50
    
Sorry, but I don't see the connection. –  xan Apr 26 '12 at 16:23

2 Answers 2

up vote 1 down vote accepted

We will use the fact that $\sum\limits_{n=1}^\infty (-1)^n a_n$ is convergent.

Keep in mind that that $(a_n)$ is a sequence of nonnegative terms that decrease to zero.

Suppose $n$ is even, say $n=2m$. Then $$\eqalign{ 0&\ge (\underbrace{-a_{2m+1}+ a_{2m+2}}_{\le 0})+(\underbrace{-a_{2m+3}+ a_{2m+4}}_{\le0})+\cdots\cr &=\sum_{k=2m+1}^\infty (-1)^k a_k \cr &=-a_{2m+1}+(\underbrace{a_{2m+2} -a_{2m+3}}_{\ge0}) +(\underbrace{a_{2m+4} -a_{2m+5}}_{\ge 0})+\cdots\cr &\ge -a_{2m+1}\cr &=-a_{n+1}. } $$

So $$\tag{1}0\ge r_n\ge -a_{n+1},\quad\text{for }\ n\ \text{even}.$$

For $n$ odd, one can show similarly that $$\tag{2}0\le r_n\le a_{n+1},\quad\text{for }\ n\ \text{odd}.$$

Inequalities $(1)$ and $(2)$ show that $|r_n|\le a_{n+1}$ and that $r_n$ has the same sign as $(-1)^{n+1}$

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Now I understand. Thank you so much. –  xan Apr 26 '12 at 16:40

We are going to use some properties and that can be found on the wikipedia. Notice that $r_n = L - S_n $ where $L = \sum_{n=0}^{\infty} (-1)^{n}a_n $ and $ S_n \sum_{n=0}^{k} (-1)^{n}a_n $. Then $$ |r_n| = | L - S_n| \le a_n, $$ and the firt term of $r_n$ is $\ge 0 $ if $n$ is odd, but if $n$ is even $$ r_n = L - S_n \ge 0 .$$ Analagously, if $n$ is even $$ r_n = L - S_n \le 0 .$$

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