convergence of subsequences of partial sums i am not sure regarding this question. could you help me?
assuming that the two sums, that are obtained from $\sum_{n=1}^\infty(a_n)$ are convergent:
$\sum_{n=1}^\infty(b_n)=(a_1+a_2)+(a_3+a_4)+...$ ,  $(b_n)=a_{2n-1}+a_{2n}$
$\sum_{n=1}^\infty(c_n)=a_1+(a_2+a_3)+(a_4+a_5)+...$, $(c_{n+1}=a_{2n}+a_{2n+1},c_1=a_1$)
I'm asked to prove or disprove the following:
1)$\sum_{n=1}^\infty(a_n)$ is convergent without any additional information.
2)if $\sum_{n=1}^\infty(b_n)=\sum_{n=1}^\infty(c_n)=S$ then $\sum_{n=1}^\infty(a_n)$ converges and is equal to S.
3)if $a_n \to 0$ then $\sum_{n=1}^\infty(a_n)$ converges and $\sum_{n=1}^\infty(a_n)=\sum_{n=1}^\infty(b_n)=\sum_{n=1}^\infty(c_n)$
my attempt:
1)$\sum_{n=1}^\infty(a_n)$ converges because the other sums derive from it, and if you change a finite number of elements from it, it will still converge to the same number. so i think it's a true claim even without any additional information.
2)since $\sum_{n=1}^\infty(b_n)$ converges, let's call the series of its partial sums S. i.e S=$\sum_{n=1}^\infty(b_n)=\sum_{n=1}^\infty(a_{2n-1}+a_{2n})$, let's do the same for the third: $S=\sum_{n=1}^\infty(c_n)=a_1+\sum_{n=2}^\infty(a_{2n}+a_{2n+1})$(because it's $c_{n+1}$ i had to work around it). so if both the sums $\sum_{n=1}^\infty(b_n)$ and $\sum_{n=1}^\infty(c_n)$ converge, and as shown they are subsequences of sums originated by changing a finite number of elements, then $\sum_{n=1}^\infty(a_n)$ must converge to the same sum S.
3)(don't know how to answer it unfortunately), but i do understand it has to do with showing the sums for $s_{2n}$ and $s_{2n+1}$ and showing their limit must be the same, so it will conclude that they converge to the same S.
thank you very much for helping me, stuck on it for a long time.
 A: Your idea is not very rigorous. These things can be made rigorous. 


*

*This is wrong. Take $a_k=(-1)^k$. Then $\sum_{k=1}^\infty a_k=\sum_{k=1}^\infty (-1)^k$ is not convergent, but $\sum_{k=1}^\infty b_k = 0$ and $\sum_{k=1}^\infty c_k=-1$. Check this yourself.

*This is true.

*This is also true.


Proof of (2). Define the following sequence:
\begin{align}
R_n:=\sum_{k=1}^\infty a_k \\
S_n:= \sum_{k=1}^\infty b_k \\
T_n:=\sum_{k=1}^\infty c_k
\end{align}
You can check the following: 
\begin{align} S_n = R_{2n}, \ \ \text{ and } \ \ T_n = R_{2n+1}\end{align}
You are given the sequence $R_n$ and you know the even subsequence and the odd subsequences converge to the same limit. So you know that the original sequence must converge to that limit as well (prove this fact yourself). So $R_n\to S$. 
Proof of (3).
You know $S_n$ and $T_n$ converge. So for every $\varepsilon>0$ there is $N$ such that for all $n>N$:
\begin{align}
\bigg|\sum_{k=n}^\infty b_k\bigg| < \varepsilon 
\end{align}
Similarly for $\sum_k c_k$. We want to prove the following:
\begin{align}
\bigg | R_n - \sum_{k=1}^\infty b_k\bigg|\to 0 \ \ \ \text{ and } \ \ \ \bigg | R_n - \sum_{k=1}^\infty c_k\bigg|\to 0
\end{align}
Let's do it. There is $N_1$ such that for all $n>N_1$ we have:
\begin{align}
\bigg|\sum_{k=n}^\infty b_k\bigg| < \frac \varepsilon 3  
\end{align}
There is $N_2$ such that for all $n>N_2$ we have $|a_{n}|<\varepsilon/3$ since $a_n\to 0$. 
So:
\begin{align}
\bigg | R_n - \sum_{k=1}^\infty b_k\bigg|=
\begin{cases}
\bigg| \sum_{k=1+n/2}^\infty b_k \bigg| & \text{ if } n \text{ is even}\\
\bigg| a_{n+1}+\sum_{k=1+(n+1)/2}^\infty b_k \bigg|& \text{ if } n \text{ is odd}
\end{cases}
\end{align}
That means:
\begin{align}
\bigg | R_n - \sum_{k=1}^\infty b_k\bigg|\leq |a_{n+1}| +\bigg|\sum_{k=1+\lfloor (n+1)/2\rfloor}^\infty b_k \bigg|+\bigg|\sum_{k=1+\lfloor n/2\rfloor}^\infty b_k \bigg|
\end{align}
Now take $N:=\max\{ 2N_1,N_2\}$, then we have for all $n>N$:
$$\bigg | R_n - \sum_{k=1}^\infty b_k\bigg|<3\frac{\varepsilon}{3} =\varepsilon$$
Thus we have shown $$\sum_{k=1}^\infty a_k= \sum_{k=1}^\infty b_k$$ 
Similarly you can show that $\sum_k a_k=\sum_k c_k$. That one is left for you.
