Show that if $s_n$ converges to $\beta$, then $t_n$ converges to $\beta/2$. Let $a_n = \frac{(-1)^n}{n},\ s_n = a_1+a_2+\cdots+a_n$
$=1-\frac12+\frac13-\frac14+\cdots+\frac{(-1)^n}{n}$
Further let 
$t_1=a_1=1$ 
$t_2=a_1+a_2=1-\frac12$ 
$t_3=a_1+a_2+a_4=1-\frac12-\frac14$ 
$t_4=a_1+a_2+a_4+a_3=1-\frac12-\frac14+\frac13$ 
$t_5=a_1+a_2+a_4+a_3+a_6=1-\frac12-\frac14+\frac13-\frac16$ 
$t_6=a_1+a_2+a_4+a_3+a_6+a_8=1-\frac12-\frac14+\frac13-\frac16-\frac18$ 
$t_7=a_1+a_2+a_4+a_3+a_6+a_8+a_5=1-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15$ 
$t_8=a_1+a_2+a_4+a_3+a_6+a_8+a_5+a_{10}=1-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\frac1{10}\\ \vdots$
that is, sum of one positive term and two negative terms.
Show that if $s_n$ converges to $\beta$, then $t_n$ converges to $\beta/2$.
I figured out that the sequence given is in the form of $\sum_{n=1}^{\infty} \bigg(\frac{1}{2n-1}-\frac{1}{4n-2}-\frac{1}{4n}\bigg)$ but I don't see how that is helpful.
Can someone give an idea as to how to start? Is there a way to use Cauchy's theorem of sequences on subsequences of $s_n$ to prove this?
Assume that series has not been defined yet, hence treating these as sequences and using the results on sequences only.
 A: The following result is known: 

If a sequence $a_n\stackrel{n\to\infty}{\longrightarrow}a$, then any subsequence of $\{a_n\}_{n\ge 1}$ also converges to $a$.

The observation you have made, if restricted to partial sums, says that $$\begin{aligned}t_{3n}&=\sum_{i=1}^n \left(\frac1{2i-1}-\frac1{4i-2}-\frac1{4i}\right)\\ &= \sum_{i=1}^n \left( \frac1{4i-2}-\frac1{4i}\right)\end{aligned}$$ so if we take the subsequence $\{t_{3n}\}_{n\ge 1}$ of $\{t_n\}$ and rename $t_{3n}$ and $w_n$, then we have $$\begin{aligned} w_n&=\sum_{i=1}^n \left( \frac1{4i-2}-\frac1{4i}\right)\\ &=\frac12-\frac14+\frac16-\frac18+\frac1{10}-\frac1{12}+\cdots-\frac1{4n}\\ &=\frac12 \left(1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots-\frac1{2n}\right)\\\implies w_n&=\frac{s_{2n}}2\end{aligned}$$
Now, $s_n\to \beta\implies s_{2n}\to \beta$ by the mentioned result, so $w_n=\dfrac{s_{2n}}2\to \dfrac{\beta}2$ 
After this, if you are willing to write $$\lim_{n\to\infty}t_n=\lim_{n\to\infty} \sum_{i=1}^\infty\left(\frac1{4i-2}-\frac1{4i}\right)=\lim_{n\to\infty}w_n=\dfrac{\beta}2$$ then you're already done.

Should you want to be more rigorous than that, given what we have proved in the answer, it should suffice to prove that the sequence $\{t_n\}$ itself actually converges so that by the abovementioned result again, any subsequence of $\{t_n\}$, in particular $\{w_n\}$ should also converge to the same limit, which we already have to be $\dfrac{\beta}2$. 
$\underline{\text{Can you prove now that the sequence $\{t_n\}$ is convergent?}}$

 This is actually a well-studied problem, thanks to Riemann and his Rearrangement Theorem (relevant upto section $3$ here), in particular, you can find a proof of what you want in this paper. What you have given is thus a rearrangement of the alternating harmonic series.

