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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.

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