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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?)?

Arctic Char
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Lucas Giraldi
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  • If $s_{2n} \to L$ and $s_{2n-1} \to L$ then $s_n \to L$. This is quite easy to prove from definition of limit. – Kavi Rama Murthy Jul 04 '20 at 07:34
  • Yeah, that's what I stated, I'm just not getting why, intuitively, this is right – Lucas Giraldi Jul 04 '20 at 07:36
  • Please show what you have done. All you need is the fact that any integer is either even or odd. – Kavi Rama Murthy Jul 04 '20 at 07:37
  • Oh, I saw the proof here: https://math.stackexchange.com/questions/187525/sufficiency-to-prove-the-convergence-of-a-sequence-using-even-and-odd-terms, but the problem is not understanding the proof and the math, but why this makes sense, you know? It's like with Leibniz theorem for series, the proof has all the maths but it does not gives you an intuitive reason for why it's true, and that's my problem here – Lucas Giraldi Jul 04 '20 at 07:41

3 Answers3

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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 $\infty$.

zkutch
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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.

tkf
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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)?

zhw.
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  • First of all, thanks a lot, marked the above one as the answer cause it answers both questions, but loved your explanation, first time that I actually understand it. – Lucas Giraldi Jul 04 '20 at 20:31
  • So, for case ii) we have $n > N$ and $n$ odd, so $n = 2(\frac{n-1}{2}) + 1$ and then $2(\frac{n-1}{2}) + 1 > N$ implies $2(\frac{n-1}{2}) + 1 > 2N_2 + 1$ and so $\frac{n-1}{2} > N_2$, which implies that $|s_n - L| = |s_{2(\frac{n-1}{2}) + 1} - L| < \varepsilon$. Right? – Lucas Giraldi Jul 04 '20 at 20:34