Home > Analysis, Problem Solving > Pseudo-alternate series

## Pseudo-alternate series

Suppose $(a_n)$ is a non-increasing sequence of positive real numbers and $\varepsilon_i \in \{\pm 1\},\ \forall i \in \mathbb{N}$ such that $\sum\limits_{i=1}^\infty \varepsilon_i a_i$ is convergent.
Prove that $\lim\limits_{n\to \infty}(\varepsilon_1+\varepsilon_2+...+\varepsilon_n) a_n=0$.

The proof for this problem was given by Fedja here. A more detailed and polished version of his answer will be posted below. There are a few interesting ideas, that might be useful in the proof of similar problems, so I’ll divide the proof in a few lemmas.

Lemma 1.If the family of sums ${ \sum_{i=k}^m b_i }$ are bounded by ${\delta}$ and ${u_i}$ is an increasing sequence of numbers in ${[1,2]}$, then the sums ${\sum_{i=k}^m b_iu_i}$ are bounded by ${2\delta}$.

Proof: The proof is easy if we use Abel’s inequality. Let ${(f_n)}$ be a sequence of positive numbers which is decreasing, and be a sequence of real(or complex) numbers. Then the following inequality holds:

$\displaystyle |\sum_{i=1}^m a_if_i| \leq A_mf_1$

where ${A_m=\max\{|a_1|,|a_1+a_2|,...,|a_1+...+a_n|\}}$. Using Abel’s inequality we have

$\displaystyle |\sum_{i=k}^m b_iu_i| \leq u_mB_m \leq 2\delta$

where ${B_m=\max\{|b_1|,|b_1+b_2|,...,|b_1+...+b_n|\}}$.

Secondly, divide the sequence ${(a_n)}$ into blocks of consecutive elements of the sequence such that if ${[a_k,...,a_p]}$ is such a block, then ${a_p \geq a_k/2 > a_{p+1}}$. Denote by ${(j)}$ the index of the first element in the ${j}$-th block and ${A_j=a_{(j)}}$. Then we can state the next result.

Lemma 2. It is enough to prove that ${\lim_{n \rightarrow \infty} A_n(\varepsilon_1+...+\varepsilon_{(n)})=0}$.

Proof: Let ${p>0}$ be a positive integer. Then there exists another unique positive integer ${j}$ such that ${a_p \in [a_{(j)},...,a_{(j+1)-1}]}$. Then we have

$\displaystyle |a_p(\varepsilon_1+..+\varepsilon_p)-A_j(\varepsilon_1+...+\varepsilon_{(j)})|=|(a_p-A_j)(\varepsilon_1+...+\varepsilon_{(j)})+a_p(\varepsilon_{(j)+1}+...\varepsilon_p)|\leq$

$\displaystyle \leq |a_p-A_j||\varepsilon_1+...+\varepsilon_{(j)}|+|a_p||\varepsilon_{(j)+1}+...+\varepsilon_p|\leq$

$\displaystyle \leq \frac{A_j}{2}|\varepsilon_1+...+\varepsilon_{(j)}|+|a_p||\varepsilon_{(j)+1}+...+\varepsilon_p|$

Choose ${\delta>0}$. Because the series ${\sum_{i=1}^\infty\varepsilon_i a_i}$ is convergent, it follows that the sums ${\sum_{i=k}^m \varepsilon_i a_i}$ can be bounded by ${\delta}$ for ${k \geq N}$ with ${N}$ large enough. Furthermore, we have

$\displaystyle a_k|\varepsilon_k+...+\varepsilon_m|=|\varepsilon_k a_k \frac{a_k}{a_k}+\varepsilon_{k+1}a_{k+1}\frac{a_k}{a_{k+1}}+...+\varepsilon_m a_m\frac{a_k}{a_m}|$

and if there exists a ${j}$ such that ${(j)\leq k\leq m \leq (j+1)-1}$ then Lemma 1 shows us that the terms

$\displaystyle a_k|\varepsilon_k+...+\varepsilon_m|$

are bounded by ${2 \delta}$. For ${j}$ large enough, if ${\lim_{n \rightarrow \infty} A_n(\varepsilon_1+...+\varepsilon_{(n)})=0}$ we have ${\frac{A_j}{2}|\varepsilon_1+...+\varepsilon_{(j)}|< \delta}$.

Then, for ${p}$ large enough, we have

$\displaystyle |a_p(\varepsilon_1+..+\varepsilon_p)|\leq |a_p(\varepsilon_1+..+\varepsilon_p)-A_j(\varepsilon_1+...+\varepsilon_{(j)})|+\frac{A_j}{2}|\varepsilon_1+...+\varepsilon_{(j)}|\leq$

$\displaystyle \leq 4\delta$

Since ${\delta>0}$ can be made arbitrarily small, Lemma 2 is now proved.

We are left to deal with the proof of the fact that ${\lim_{n \rightarrow \infty}A_n(\varepsilon_1+...+\varepsilon_{(n)})=0}$. First, let’s notice that we have

$\displaystyle |A_{n+1}||\varepsilon_1+...+\varepsilon_{(n+1)}|\leq \frac{|A_n|}{2}|\varepsilon_1+...+\varepsilon_{(n)}|+|A_{n+1}||\varepsilon_{(n)+1}+...+\varepsilon_{(n+1)}|$

and by using the same trick as in the proof of Lemma 2, we see that

$\displaystyle |\varepsilon_{(n)+1}+...+\varepsilon_{(n+1)}|\rightarrow 0$

as ${n \rightarrow \infty}$. Denote

$\displaystyle x_n=|A_n(\varepsilon_1+...+\varepsilon_{(n)})| \text{ and }y_n=|A_{n+1}||\varepsilon_{(n)+1}+...+\varepsilon_{(n+1)}|.$

We know that ${0 \leq x_{n+1}\leq \displaystyle \frac{x_n}{2}+y_n}$ and ${y_n \rightarrow 0}$. Take ${\limsup}$ in the previous inequality and get

$\displaystyle 0 \leq \limsup x_n \leq \frac{1}{2} \limsup x_n.$

This implies that ${\limsup x_n=0}$. Since ${x_n \geq 0}$, we find that ${\displaystyle\lim_{n\rightarrow \infty}x_n=0}$ and that finishes the proof.