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What are the argument(s) that I can use proving that

$$1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots)$$

does not exist.

The question was:

Find a arrangement of $\sum\frac{(-1)^{n-1}}{n}$ for which the new sum is not exist(even not $+\infty$ or $-\infty$)

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    $\begingroup$ This is a confusing definition of the word "sum." You have two infinite series. $\endgroup$ May 31, 2013 at 16:31
  • $\begingroup$ both term diverges to $\infty$ $\endgroup$
    – hasExams
    May 31, 2013 at 16:33
  • $\begingroup$ The rules for operator precedence might be invoked. In particular the "expression" inside the parentheses needs to be evaluated first. However that "expression" appears to be a divergent infinite series, so one can argue its evaluation is not possible. $\endgroup$
    – hardmath
    May 31, 2013 at 16:34
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    $\begingroup$ What you have is not a valid rearrangement. $\endgroup$
    – Potato
    May 31, 2013 at 16:37
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    $\begingroup$ For the original question you were asked, take positive terms until the sum is $\ge 1$, then negative terms until the sum is $\le 0$, then positive terms until the sum is $\ge 1$, then negatives until sum is $\le 0$, and so on. The first few are $1-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}+\frac{1}{3}+\frac{1}{5}+ \cdots $. $\endgroup$ May 31, 2013 at 16:43

4 Answers 4

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One should be careful in dealting with infinity series.

The sum $$ \sum_{n=1}^\infty(-1)^{n-1}\frac 1 n $$ is well known as an alternating series, which converges(to $\ln 2$).

However, the series $$ \left(1+\frac 1 3 + \frac 1 5 +\ldots \right)-\left(\frac 1 2 + \frac 1 4 +\ldots \right) $$ does not exist if not specitied. The series $$ \lim_{n\to \infty}\left(\sum_{k=1}^n\frac 1 {2k-1}-\sum_{k=1}^n\frac 1 {2k}\right) $$ exists and is $\ln 2$.

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Here is a hint for your problem

$$1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots$$

diverges. So does $$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots$$

Lets denote the partial sums by

$$S_n = \sum_{k=1}^n \frac{1}{2k-1} \,;\, T_n = \sum_{k=1}^n \frac{1}{2k} \,.$$

Now pick $n_1$ so that

$$S_{n_{1}} >1 \,.$$

Pick $m_1$ so that $T_{m_1}> S_{n_1}$.

Pick inductively $n_i, m_i$ so that

$$S_{n_{i}} >1+T_{n_{i-1}} \,,$$ $$T_{m_i}> S_{n_1} \,.$$

Then

$$S_{n_1}-T_{n_1}+(S_{n_2}-S_{n_1})-(T_{n_2}-T_{n_1})+(S_{n_3}-S_{n_2})-(T_{n_3}-T_{n_2})+...$$

oscillates above 1 and below 0. More exactly, for all $k$,

$$S_{n_1}-T_{n_1}+(S_{n_2}-S_{n_1})-(T_{n_2}-T_{n_1})+(S_{n_3}-S_{n_2})-(T_{n_3}-T_{n_2})+...+(S_{n_k}-S_{n_{k-1}})>1\,,$$ $$S_{n_1}-T_{n_1}+(S_{n_2}-S_{n_1})-(T_{n_2}-T_{n_1})+(S_{n_3}-S_{n_2})-(T_{n_3}-T_{n_2})+...+(S_{n_k}-S_{n_{k-1}})-(T_{n_k}-T_{n_{k-1}})<0\,,$$

P.S. The idea of the proof is simple to understand: add enough positive terms to go over 1. Then subtract enough negative terms to go under 0. Then add enough of the next positive terms to go over 1. Then subtract enough of the next negative terms to go under 0. Repeat.

You can do this process because of the above series go to $\infty$.

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  • $\begingroup$ Sir there is some latex error . $\endgroup$ Jan 15, 2018 at 8:36
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This does exist. The series is $$\sum\limits_{i=1}^\infty (-1)^{i+1} a_i$$ where $a_i=\frac{1}{i}$ which is an alternating seres. Since the limit of $a_i$ as $i$ tends to infinity is zero and $a_i$ is monotonically decreasing, the series converges.

Although, you should be a bit careful writing things this way. The two series should really be merged so it doesn't look like you're trying to do arithmetic with infinities.

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  • $\begingroup$ The limit of terms $a_i$ being zero is necessary, but not sufficient, for the series to converge. $\endgroup$
    – hardmath
    May 31, 2013 at 16:38
  • $\begingroup$ true, the $a_i$ should also be monotonically decreasing which I've added in $\endgroup$
    – john
    May 31, 2013 at 16:42
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    $\begingroup$ Thes series you mention is not equal to the OP's $\sum_{k\geq 0}\frac{1}{2k+1}-\sum_{k\geq 1}\frac{1}{2k}$ which, indeed, does not define a real number, not even $\pm \infty$. $\endgroup$
    – Julien
    May 31, 2013 at 16:55
  • $\begingroup$ technically no in the way it is written. Hence the last paragraph. $\endgroup$
    – john
    May 31, 2013 at 17:08
  • $\begingroup$ I'm not sure your answer is of much help for the OP, but note that I did not downvote as everything you say is true. $\endgroup$
    – Julien
    May 31, 2013 at 17:22
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Take 2 positive terms, 4 negative terms, 8 positive, 16 negative, and so on. The resulting series of summations will oscillate between positive and negative values, never approaching any limit.

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