Divergent series : its subsequence sum to arbitrary number

Question

Let $(a_n)$ be a sequence of positive real number. Assume that

$$\sum_{n=1}^\infty a_n = \infty, \lim_{n \rightarrow \infty} a_n = 0.$$ Then for any positive real number $a$, there exists a subsequence $(a_{n_k})$ such that $$\sum_{k =1}^\infty a_{n_k} = a.$$

The statement is, more or less, analogous to Riemann rearrangement theorem for conditionally convergent series (http://www.cut-the-knot.org/arithmetic/algebra/RiemannRearrangementTheorem.shtml). I can imagine why the theorem is possible. It is like $a_n$ each is tiny, but when it sum together it can be gigantic. So for any positive number $a$, it is like I need to have a finite sum of $a_n$ up to where it is near $a$. Then use the tail term of $a_n$ that really small, say might be smaller than $1/2^n$ so it infinite sum is tiny, to fill up the gap between to finite sum of $a_n$ and $a$.

However, I cannot write the proof. It is hard to set up the finite sum until it almost reach $a$. I try to mimic the proof in Riemann theorem, but it is considerably different form this statement.

Help please.

Answer 1

Construct the sequence by hand (that's actually the proof that I know from Riemann's theorem):

Given $n_1, \cdots , n_k $, we pick $n_{k+1} $ as the smallest index $i > n_k$ such that $a_i < a - \sum\limits_{j=1}^k a_{n_j}$.

This process converges because the tails always sum $\infty $.

Answer 2

Claim: There are positive integers

$$m_1 \le n_1 < m_2 \le n_2 < m_3 \le n_3 < \cdots $$

such that if $S_k = \sum_{n=m_k}^{n_k}a_n,$ then

$$\tag 1 a-1/k < S_1 + \cdots + S_k <a$$

for all $k.$ Supposing the claim is true, set $A = \cup_k \{n: m_k \le n \le n_k\}.$ Then the desired subsequential series is $\sum_{n\in A} a_n.$

Proof of claim: Suppose $m_1,n_1, \dots , m_k,n_k$ have been chosen as above so that $(1)$ holds. Then there exists $m_{k+1}>n_k$ such that $n\ge m_{k+1}$ implies

$$a_n < \min(a-(S_1+ \cdots +S_k), 1/(k+1)).$$

Note that $S_1+ \cdots +S_k + a_{m_{k+1}} < a.$ Because $\sum_{m_{k+1}}^{\infty}a_n = \infty,$ there exists a smallest $N> m_{k+1}$ such that

$$S_1+\cdots + S_k +\sum_{m_{k+1}}^{N}a_n \ge a.$$

Define $n_{k+1} = N-1.$ Let's check that $(1)$ holds with $k+1$ in place of $k.$ That $S_1+\cdots + S_{k+1}<a$ is easy. For the lower bound we have

$$ S_1+\cdots + S_{k+1} + a_N \ge a \implies S_1+\cdots + S_{k+1} \ge a - a_N > a-1/(k+1).$$

This proves the claim except for a small point: I didn't deal with $S_1$ at the beginning. It should be clear how that would go.

Answer 3

Your claim easily follows from the following lemma:

Lemma: let $(a_n)_n$ be a sequence of positive numbers converging to zero such that $\sum_na_n=\infty$, and let $c$ be a positive real number. Then there exist some $n_1,n_2$ such that $$\frac c 2<\sum_{n=n_1}^{n_2}a_n<c$$

(To derive you claim, recursively plug in appropriate tails of the original sequence and the remaining part of $a$.)

To prove the lemma: let $n_1$ be such that for all $n\geq n_1$ we have $a_n<\frac c 2$, and let $n_2$ be the smallest integer such that $\sum_{n=n_1}^{n_2} a_n>\frac c 2$ (this exists because $\sum_{n=n_1}^\infty a_n=\infty$). Since $a_{n_2}<\frac c 2$ and $\sum_{n=n_1}^{n_2-1}a_n\leq \frac c 2$, we have $\sum_{n=n_1}^{n_2} a_n<\frac c 2+\frac c 2=c$. $\square$