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Test the convergence of summation $$\sum_{n=1}^\infty x_n$$ where $$x_{2n-1}=\frac{n}{n+1}\\ x_{2n}=-\frac{n}{n+1}$$ That is the series $$\frac 1 2-\frac 12+\frac 23-\frac 23 +-\cdots$$

what I did was let Sn be the partial sums of the series.Then

$$S_n=\begin{cases} 0 & \text{when } n \text{ is even} \\ \frac{n}{n+1} &\text{when } n \text{ is odd}\end{cases}$$

thus $$\lim\limits_{ n\to \infty} S_n= \begin{cases} 0 & \text{when } n \text{ is even}\\ 1 & \text{when } n \text{ is odd}\end{cases}$$

Thus $\lim\limits_{ n\to \infty} S_n$ doesn't converge to a particular value. Hence $\lim\limits_{ n\to \infty} S_n$ doesn't exist. Therefore the series diverge.

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    $\begingroup$ Please see here for how to typeset common math expressions with MathJax, and see here for how to use Markdown formatting. $\endgroup$ – Zev Chonoles Jul 1 '13 at 17:54
  • $\begingroup$ Your formula for $S_n$ when $n$ is odd is wrong. The formula for $S_{2n-1} =\frac{n}{n+1}$. $\endgroup$ – Thomas Andrews Jul 1 '13 at 18:09
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Your answer is correct, but I would say it differently. What you mean is that $$\lim_{n\to\infty}S_{2n}=0$$ $$\lim_{n\to\infty}S_{2n+1}=1$$

Note that if we "introduced parenthesis", and set $$a_n=x_{2n-1}+x_{2n}$$ the series $$\sum a_n$$ would converge to $0$.

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  • $\begingroup$ What you mean by introduced paranthesis is after a rearrangement of the series right?I want to know if my statement "Thus limn→∞Sn doesn't converge to a particular value. Hence limn→∞Sn doesn't exist. Therefore the series diverge." is correct. $\endgroup$ – clarkson Jul 2 '13 at 18:49
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A much easier way, imo, to show the series doesn't converge:

$$|x_n|=\frac n{n+1}\xrightarrow[n\to\infty]{}1\neq 0$$

and thus the series $\,\sum x_n\,$ cannot converge.

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  • $\begingroup$ Is it okay to consider only limit of n/n+1.Because clearly this is not $a_n$ of the series.This was done by divergence test right? (since limit of $a_n$ is not 0 the series diverge).Also if we have a series which is made up of two series where one converge and the other diverge, does the complete series diverge $\endgroup$ – clarkson Jul 2 '13 at 18:53
  • $\begingroup$ Yes, it is o.k. since even if it is only a subsequence the fact that its limit isn't zero shows that the general term sequence doesn't converge to zero. $\endgroup$ – DonAntonio Jul 2 '13 at 21:19
  • $\begingroup$ This is a series not a sequence.For a sequence I know that every subsequence should converge.But does the same thing apply for a series? $\endgroup$ – clarkson Jul 3 '13 at 2:34
  • $\begingroup$ A series is just a sequence of partial sums. $\endgroup$ – Paul Malinowski May 30 '14 at 16:10
  • $\begingroup$ No Paul: a series is not a sequence, neither of partial sums or anything else. In order to find out whether a series converges or not then we take its partial sums' sequence, but those are two rather different things. $\endgroup$ – DonAntonio May 30 '14 at 16:25

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