The sum $1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots)$ does not exist. 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 $-\infty$) 
 A: 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 $\ln 2$.
A: 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 $\infty$. 
A: 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.
A: 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.
