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.

Solution 1:

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 $\ln 2$.

Solution 2:

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 $\infty$.