Convergence of subseries and partitioning the natural numbers I'm learning analysis at this semester and I stumbled upon a theorem that says:

Let $(a_ n)_{n \in \mathbb{N}}$ be a sequence and $s_n = \displaystyle\sum_{k = 1}^{n} a_n$ the sequence of partial sums of the series $\displaystyle \sum_{k = 1}^{\infty} a_n$. If $\displaystyle\lim_{n \to \infty} s_{2n} = L$ and $\displaystyle\lim_{n \to \infty} s_{2n + 1} = L$, then $\displaystyle \sum_{k = 1}^{\infty} a_n = L$.

My first question is simple: how does one proves this? I'm trying to understand why this is true but it just does not seems to fit.
Now, the interesting question:

Given $\mathcal{P}$ a partition of $\mathbb{N}$, if $\displaystyle\lim_{n \to \infty} s_{p_n} = L$, where $p_n$ are the elements of some set of $\mathcal{P}$, and this happens for all sets of $\mathcal{P}$, then $\displaystyle \sum_{k = 1}^{\infty} a_n = L$.

Is this true? I mean, can one gives an intuitive reason why it isn't (yeah counter-examples work but I really want to feel it in my veins, I kind want it to make sense, you know?)?
 A: Let's take $N_1 \cup N_2 = \mathbb{N},N_1 \cap N_2 = \emptyset$, both are infinite. In first case $N_1$ are odd and $N_2$ are even natural numbers.
Let's consider any sub sequence $s_{n_{k}}$ of sequence $s_n$. There are 3 cases: all, but finite part of $s_{n_{k}}$ members lies in 1)$N_1$ 2)$N_2$ 3) in both $N_1$ and $N_2$. In first and second cases $s_{n_{k}}$ has limit and in third case it have, generally 2 limit points, but as these limit points are same, then again it converges.
General case, assuming we have finite partition of $\mathbb{N}$ to $k$ patrs, is repeating above for $k$ patrs.
Addition: Suppose we have infinite partition i.e. $\mathbb{N} = \cup_{i=1}^{\infty}N_i$, for $i \ne j$ we have $N_i \cap N_j = \emptyset$ and all $\forall i, N_i$ are infinite and exists same limit $L$ against it.
Now let's change values for all $s_n$ for first members of $N_i$ and let's set it to $i$. If we now consider sub sequence, which consist from first members of $N_i$ $\forall i \in \mathbb{N}$, then it converges to $\infty$.
A: Let $P_1=\{1\} \cup T_1,P_2=\{3\}\cup T_2,P_3=\{5\}\cup T_3\cdots$, where $T_i$ consists of all natural numbers divisible by $2^i$, but not $2^{i+1}$.
Let $a_i=(-1)^i$.  Then the $s_i|i\in P_j$ converge to $0$ for any $j$.  However  the sum over $\mathbb{N}$ does not converge.
A: This can be a little weird the first time you go though it. It's almost like a change of variables.
Take any real sequence $s_n$ and let $L\in \mathbb R.$ Suppose $\lim_{n\to \infty} s_{2n}=L$ and $\lim_{n\to \infty} s_{2n+1}=L.$ We want to show $\lim_{n\to \infty} s_{n}=L.$
So let $\epsilon>0.$ Then there exists $N_1$ such that $n>N_1$ implies $|s_{2n}-L|<\epsilon.$ And there exists $N_2$ such that $n>N_2$ implies $|s_{2n+1}-L|<\epsilon.$
Set $N=\max (2N_1,2N_2+1).$ Let $n>N.$ Two cases: i) $n$ is even ii) $n$ is odd. In the first case, we write $n=2(n/2).$ Then $2(n/2)>N$ implies $2(n/2)>2N_1,$ which gives $n/2 >N_1.$ Therefore
$$|s_n-L|= |s_{2(n/2)}-L|<\epsilon.$$
Would you like to try your hand at case ii)?
