Let $\{a_n\}$ be a sequence of positive real numbers such that the sum $\sum a_n$ converges. What condition is necessary and sufficient to ensure that, for any positive real number $r$, less than $\sum a_n$, there is a subsequence $\{a_{n_k}\}$ such that $\sum a_{n_k} $ converges to $r$?

I have found what I think to be a necessary condition, but I am unsure if it alone is sufficient.

(Possible) Solution via contrapositive:

Given the above, assume there exists some positive $r<a$ such that no subsequence $A_k=\{a_{n_k}\}$ exists, whose infinite sum $\sum a_{n_k}$ converges to $r$. For ease, let $S_k=\{s_{n_k}\}$ be the sequence of partial sums of some $A_k$ and $S=\{S_k\}$ the set of all sequences of partial sums of the $A_k$. Thus, stated in these terms, it follows there is no $s \in S$, such that $$s \to r.$$ This means there exists a real number $p>0$, such that, if $x \in N_p(r)$, then $x\neq a_{m_1} +a_{m_2}+\cdots +a_{m_n}$.

(Attempted) Explicit Construction:

Assume, now, that for any $r<a$ and any $\epsilon_m>0$, there exists a finite sum of elements $s_{l_m}=\displaystyle \sum_{k=l_1}^{l_m}a_k$ such that $s_{l_m} \in (r -\epsilon_m, r)$.

Given this, and since $a_n \to 0$, then, for any $0<\epsilon_{m+1}<\min \{a_k ,\epsilon_m\}$, we can find $a_{l_{m+1}}$ such that

$$a_ {l_{m+1}} < \epsilon_{m+1}<\epsilon_m.$$

Thus, we get $$r- \epsilon_m < s_{l_m} < s_{l_m} + a_ {l_{m+1}} <s_{l_m} + \epsilon_{m+1}<r.$$

Next, being careful with our selection of $\epsilon_{m+2}$, $$r- \epsilon_m <s_{l_m} + a_ {l_{m+1}} + a_ {l_{m+2}} <r.$$

We can continue this process indefinitely. The sequence $\{s_ {l_{m+n}}\}$ of partial sums defined by $$s_ {l_{m+n}} =\displaystyle \sum_{k=l_1}^{l_{m+n}}a_k$$ is increasing and bounded above by $r$, and so it converges to some number $l\leq r$. If $l=r$, we are done. If $l<r$, then there exists a subsequence $\{a_{n_m}\} $ of $S$ such that:

$$r- \epsilon_m <\sum a_{n_m} < r.$$

At this point, it seems tempting to say, since $\epsilon_m$ was chosen arbitrarily, the result follows, but since each choice of $\epsilon_m$ give a unique finite sum $s_ {l_{m+n}} =\displaystyle \sum_{k=l_1}^{l_{m+n}}a_k$, this doesn’t seem like a simple limiting process.

Could I use the theorem which states that the set of subsequential limits of a sequence $\{p_n\}$ form a closed subset? Or is this line of reasoning completely off track?

  • 1
    $\begingroup$ It certainly works for $\{2^{-n}\}_{n=1}^{\infty}$, and does not work for $\{2^{-n^2}\}_{n=1}^{\infty}$. $\endgroup$ – Michael Jun 20 '18 at 4:17
  • $\begingroup$ @Michael is the “It” in that sentence the condition that there must exist a finite sum of $a_i$ in every interval $(r-\epsilon, r)$? $\endgroup$ – Moed Pol Bollo Jun 20 '18 at 5:52
  • $\begingroup$ I mean that it allows all numbers $r \in [0, \sum_{i=1}^{\infty} a_i]$ to be achieved over subsequence sums. Because any number $r \in [0,1]$ can be written in a binary expansion $r=\sum_{i=1}^{\infty} b_i2^{-i}$ with $b_i \in \{0,1\}$ for all $i$. The answer below gives the general condition. $\endgroup$ – Michael Jun 20 '18 at 15:58

Assuming that $(a_n)$ is a decreasing positive summable sequence, the following equivalence holds : you can write all positive reals less than $\sum a_n$ as sums of subsequences of $(a_n)$ if and only if for all $n$, $a_n \le \sum\limits_{k > n} a_k$.

$ $

Remark: your problem essentially boils down to this one (where we assume $(a_n)$ is decreasing), because as the sequence is summable, we can reorder it as we please without changing the sums.

Proof: let us assume that there exists $n$ such that $a_n > \sum\limits_{k>n}$. Consider $B \subset \mathbb{N}$. If $B$ contains an integer $\le n$, then $\sum\limits_{k\in B} a_k \ge a_n$, and otherwise, $\sum\limits_{k\in B} \le \sum\limits_{k>n}a_k$. Hence no real in $\big]\sum\limits_{k>n}a_k,\ a_n\big[$ can be written as the sum of a subsequence of $(a_n)$.

$ $

Conversely, assume that for all $n$, $a_n \le \sum\limits_{k>n}a_k$. Take $x \in \big[0,\sum\limits_k a_k\big]$. Define $n_0 := \min \{k\in\mathbb{N} \ |\ a_k < x\}$, and for all $p \ge 0$, $n_{p+1}:=\min\{k>n_p\ |\ \sum\limits_{i=0}^p a_{n_i} + a_k < x\}$. We prove by induction that for any $p \ge 0$, $x - \sum\limits_{i > n_{p+1}} a_i \le \sum\limits_{i=0}^p a_{n_i} < x$.

  • Base case if $n_0=0$, we are just saying that $x \le \sum\limits_{i\ge 0}a_i$. Otherwise, the definition gives us $x \le a_{n_0-1}$, so using our hypothesis, $x \le \sum\limits_{i\ge n_0}a_i$, and we are done.

  • Inductive step assume it to hold for some $p\ge 0$. The right inequality obviously holds. Then, if $n_{p+1}=n_p+1$, the right inequality is the same, but just adding $a_{n_{p+1}}$. Otherwise, the definition gives us $x \le \sum\limits_{i=0}^p a_{n_i} + a_{n_{p+1}-1}$, and thus $x \le \sum\limits_{i=0}^p a_{n_i} + \sum\limits_{j\ge n_{p+1}}a_j$, which is the right inequality.

Finally, as $n_p \underset{p\to\infty}{\longrightarrow}\infty$, the lower bound goes to $x$ and thus $\sum\limits_i a_{n_i}=x$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.