Let $s_k$ be the $k$th partial sum of $\sum x_n$.

a) Suppose that there exists a $m \in \mathbb{N}$ such that $\lim_{k\to\infty} s_{mk}$ exists and $\lim x_n = 0$. Show that $\sum x_n$ converges.

b) Find an example where $\lim_{k \to \infty} s_{2k}$ exists and $\lim x_n \not=0$ (and therefore $\sum x_n$ diverges).

c) (Challenging) Find an example where $\lim x_n =0$, and there exists a subsequence $\{s_{k_j}\}$ such that $\lim_{j \to \infty} s_{k_j}$ exists, but $\sum x_n$ still diverges.

I know that the series converges iff $s_k$ converges. So, I think that the question a) simply follows from this equivalence (am I right?). For b), I think that this series should be the one that is alternating and each term cancel out only when the series is even partial sums, but I can't find such series. I have no idea for c).

This question is a bit challenging to me. I appreciate if you give some help.

  • $\begingroup$ Your observation on convergence of $s_k$ being equivalent to convergence of $\sum_{n=1}^{\infty} x_n$ seems to be the definition of convergence (not really an "equivalence" that will help). For (a), you don't know if convergence holds, you must prove it using the fact that conergence holds when we consider the sum in blocks of size $m$. $\endgroup$ – Michael Dec 16 '19 at 23:07
  • $\begingroup$ For (b) I am baffled why you say you cannot find such a series when you have already described it. $\endgroup$ – Michael Dec 16 '19 at 23:09
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    $\begingroup$ For b) we can just take $x_{2n}=-1$, $x_{2n+1}=1$. Then (assuming we are starting the sum at zero) $s_{2n}=-1$ for all $n$, so clearly $\lim_{n\to\infty} s_{2n}=-1$. But $$\limsup_{n\to\infty}x_n = 1 > -1 = \liminf_{n\to\infty} x_n$$ so $x_n\not\stackrel{n\to\infty}\longrightarrow 0$. $\endgroup$ – Math1000 Dec 16 '19 at 23:10

For $a):$ Given $\epsilon >0,$ take $k_1\in \Bbb N$ such that $k_1< k\le k'\implies |s_{km}-s_{k'm}|<\epsilon/3.$ Take $k_2\in \Bbb N$ with $k_2\ge k_1$ and such that $k_2m< j\implies |x_j|<\epsilon/(3m).$

Now for $k_2m\le n<n',$ consider $k,k'\in \Bbb N$ where $km>n\ge (k-1)m$ and $k'm>n'\ge (k'-1)m.$

We have $k_1< k\le k'$ so $$|s_{km}-s_{k'm}|<\epsilon/3.$$ We have $k_2m< n+1 \le km $ so $$|s_n-s_{km}|=|\sum_{j=n+1}^{km}x_j\,|\le \sum_{j=n+1}^{km}|x_j|<$$ $$<(km-n)\cdot \epsilon/(3m)\le m\cdot \epsilon/(3m)=\epsilon/3.$$ Similarly we have $$|s_{k'm}-s_{n'}|<\epsilon /3.$$

So $|s_n-s_{n'}|\le |s_n-s_{km}|+|s_{km}-s_{k'm}|+|s_{k'm}-s_{n'}|<\epsilon.$

For $c):$. Take a sequence $(b_n)_{n\in \Bbb N}$ of members of $(0,1]$ such that $\lim_{n\to \infty}b_n=0$ and $\sum_{n\in \Bbb N}b_n=\infty.$ E.g. $b_n=1/n.$

Let $a_1=b_1.$

If $s_n\ge 1$ then $a_{n+1}=-b_{n+1}.$

If $s_n\le 0$ then $a_{n+1}=b_{n+1}.$

If $s_n\in (0,1)$ and $a_n>0$ then $a_{n+1}=b_{n+1}.$

If $s_n\in (0,1)$ and $a_n<0$ then $a_{n+1}=-b_{n+1}.$

For any $r\in [0,1],$ the sequence $S=(s_n)_{n\in \Bbb N}$ has a subsequence converging to $r.$ For your Q, it suffices to show that $S$ has a subsequence converging to $1$ and another subsequence converging to $0.$

$S$ is like a snail that wanders back and forth from approximately $1$ to approximately $0$ in "eventually smaller" steps.

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  • $\begingroup$ A particular example for $c)$: Let $x_1=+1,\;$ $ x_2=x_3=-1/2,\;$ $x_4=x_5=x_6=+1/3,\; $ $x_7=x_8=x_9=x_{10}=-1/4,\;$ $ x_j=+1/5$ for $11\le j\le 15,$ et cetera, with infinitely many $n$ for which $s_n=0,$ and infinitely many $m$ for which $s_m=1.$ $\endgroup$ – DanielWainfleet Dec 17 '19 at 13:14

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