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Use Cauchy's convergence criterion to prove convergence of $x_n$


as far as I am concerned,supposing that $m>n$, it's apparently warranted that $|x_m-x_n|=\sum_{k=n+1}^m (-1)^{k+1}\dfrac{1}{k}$

but, furthermore, it was showed on the textbook that $\sum_{k=n+1}^m (-1)^{k+1}\dfrac{1}{k}<\dfrac{1}{n+1}$

I wonder how can we approach the last step

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This may help. – David Mitra Jun 14 '12 at 17:36
up vote 2 down vote accepted

My post will going to aim and clarify that inequality in your textbook by using some elementary approach. First of all, notice that each term in your series that is set on an odd position is positive. Let's fix n as an even number and fix an even number of terms of your series and then we get:

$$S=\sum_{k=n+1}^m (-1)^{k+1}\dfrac{1}{k} = \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3}-\frac{1}{n+4}+\frac{1}{n+5}-\frac{1}{n+6}+\cdots $$ $$= \frac{1}{(n+1)(n+2)}+\frac{1}{(n+3)(n+4)}+\frac{1}{(n+5)(n+6)}+ ...$$

Here we are going to use an inequality with a sum that telescopes (on the right side):

$$S< \frac{1}{(n+1)(n+2)}+\frac{1}{(n+2)(n+3)}+\frac{1}{(n+3)(n+4)}+ ...=$$ $$\frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+2}-\frac{1}{n+3}+\frac{1}{n+3}-\frac{1}{n+4}+...<\frac{1}{n+1}.$$ Notice that if you have an odd number of terms, then when you write down the inequality and
take out from both sides the term $\frac1{n+1}$ then on the left side you get an even number of terms and you can group them such that every pair is a negative sum. Moreover, if your sum starts with a negative number, then the sum is obviously negative because the terms of the sum decrease. Therefore, the inequality holds.

The proof is complete.

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@ Chris : When the sum starts with a négative number , the sum $S$ is negative then : $S \leq \frac 1{n+1}$. Ok , but the thing Cauchy criterion needs is not $S < \varepsilon$ but $|S| < \varepsilon$. – Mohamed Jun 15 '12 at 1:13
@Mohamed: I tried to wash away his doubt regarding the reson for that in his textbook $\sum_{k=n+1}^m (-1)^{k+1}\dfrac{1}{k}<\dfrac{1}{n+1}$ appears. Read the beginning of my comment. Thanks. – user 1618033 Jun 15 '12 at 7:36
thanks, I have got the point.For upper,I think that we can reach the requirement for $|S|<\varepsilon$ by the similar method when we have sum started with a positive number.when the sum start with positive number,$|\sum_{k=n+1}^{m}-1^{k+1}\dfrac{1}{k}|=|-\dfrac{1}{n+1}+\dfrac{1}{n+2}-.‌​...|=|-\dfrac{1}{(n+1)(n+2)}-\dfrac{1}{(n+3)(n+4)}....|=\dfrac{1}{(n+1)(n+2)}+\df‌​rac{1}{(n+3)(n+4)}....<\dfrac{1}{n+1}$ thanks for help – Francis King Jun 15 '12 at 7:51
@Francis King: if you factorize $(-1)$ things are even simpler. Welcome. – user 1618033 Jun 15 '12 at 7:56

Well, Cauchy's Convergence Theorem tells that the series $\,\displaystyle{\sum_{n=1}^\infty a_n}\,$ converges iff $$\forall\,\epsilon>0\,\,\exists M\in\mathbb{N}\,\,s.t.\,\,n,m>M\Longrightarrow |S_m-S_n|<\epsilon\,\,,\,\,with\,\,S_k:=\sum_{n=1}^ka_n$$or on other words: iff the sequence of partial sums of the series is a Cauchy sequence.

In your case, it is enough for you to choose $$\,n\in\mathbb{N}\,\,s.t.\,\,\frac{1}{n+1}<\epsilon$$to prove the convergence of your sequence...

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You do not have all times : $|x_m-x_n| = \displaystyle \sum_{k=n+1}^m \frac{(-1)^{k+1}}{k}$: this can be negative

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For example , take $n=1$ and $m=4$, then : $$\displaystyle \sum_{k=2}^4 \frac{(-1)^{k+1}}{k}=-\frac 12+\frac 13 - \frac 14 = - \frac 5{12} < 0 $$ – Mohamed Jun 14 '12 at 18:24

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