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let $a_n$ be a series and +5, -5 partial limits of it.

as well note that $\left|a_{n+1}-a_{n}\right| \leq \frac{1}{n}$ for all $n$.

prove that 0 is also a partial limit of $a_n$.

Thank you.

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"As well note that" implies that we already can prove $|a_{n+1}-a_n|\le\frac{1}{n}$ from what we know already but you are really giving us more information about the series. For example, the series $5, -5, 5, -5, 5, -5, \dots$ has partial limits $5$ and $-5% but does not satisfy the second sentence. –  Ross Millikan Feb 26 '11 at 15:30
    
Thank you, I'll take it to my attention. –  user6163 Feb 26 '11 at 16:17

2 Answers 2

up vote 3 down vote accepted

It might be easiest to do a proof by contradiction.. suppose $0$ is not a partial limit. Then there is an $\epsilon > 0$ such that $|a_n| > \epsilon$ for all $n$. So if $n > {1 \over \epsilon}$, $a_{n+1}$ must have the same sign as $a_n$ since $|a_{n+1} - a_n| \leq {1 \over n} < \epsilon$. As a result, all $a_n$ for $n > {1 \over \epsilon}$ have the same sign, preventing both a positive and negative number from being a partial limit.

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Can You please more Clarify why an+1 must have the same sign as an ? –  user6163 Feb 26 '11 at 16:26
    
I'm just saying if $a_n > 0$ for some $n > {1 \over \epsilon}$, then $a_m > 0$ for $m > n$, and if $a_n < 0$ for some $n > {1 \over \epsilon}$, then $a_m < 0$ for $m > n$. If the first case holds, all partial limits are nonnegative (so $-5$ can't be), and if the second case holds all partial limits are nonpositive (so $5$ can't be). –  Zarrax Feb 26 '11 at 16:51

The idea is this:

Imagine a child draws the real line on the ground and then starts hopping around on it in such a way that again and again, they land very close to "5" and to "-5". They can do this without ever landing close to "0": for example, they can just hop back and forth between "5" and "-5".

But now imagine that as time goes on, the child gets tired and their hops become smaller and smaller (actually: arbitrarily small). Then if they still go back and forth between "5" and "-5", then on the way they'll eventually have to land on or very close to "0": once their hops get smaller than $\epsilon$, if you divide the interval from "5" to "-5" into boxes of size $\epsilon$, then on their way from the one to the other, they will have to land in every box, including the box containing "0", since their hops aren't large enough to jump over an entire box in one hop. And if you wait long enough, you can do this for arbitrarily small $\epsilon$.

Can you turn that little story into a formal proof? :-)

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+1: a nice way of thinking about it, and it leaves something for the OP to do. –  Pete L. Clark Feb 26 '11 at 17:11

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