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This is an old qual problem I'm working on: Let $\{a_k\}$ and $\{b_k\}$ be two infinite sequences of real numbers. Suppose that $a_k>0$ and $0\leq b_k \leq 1$ for all $k\geq 1$, and suppose that $\sum_{k=1}^{\infty} a_k= +\infty$. Prove that there is an increasing sequence $\{k_n\}$ of positive integers such that

i) $\sum_{n=1}^{\infty} a_{k_n}= +\infty$

ii) $\lim_{n\rightarrow \infty} b_{k_n}$ exists.

I tried to show that $\{a_n\}$ has a divergent subsequence whose all subsequences are again divergent. Then, clearly this will finish the problem, because any subsequence of ${b_n}$ has a convergent subsequence, by the compactness of $[0,1]$. However, what I tried to show might be very false too, I'm not sure. I would appreciate any help. Thanks!

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    $\begingroup$ You meant $\sum_{n=1}^\infty$ in (i). $\endgroup$ – David C. Ullrich Aug 13 '15 at 16:11
  • $\begingroup$ Yes, sorry. I'm editing now. $\endgroup$ – vgmath Aug 13 '15 at 16:12
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I'm going to assume $0\le b_k<1$, just because half-open intervals are nicer (for example the two intervals $[0,1/2)$ and $[1/2,1)$ form a partition of $[0,1)$, while we cannot write $[0,1]$ as the union of two disjoint closed intervals).

Given half-open interval $I$, say $I_L$ and $I_R$ are the right half and left half of $I$.

Let $I_0=[0,1)$. Let $$A_0=\{k:b_k\in(I_0)_L\},\quad B_0=\{k:b_k\in(I_0)_R\}.$$ One of $\sum_{k\in A_0}a_k$ and $\sum_{k\in B_0}a_k$ must be infinite; suppose wlog $\sum_{k\in A_0}a_k$ is infinite. Let $I_1=(I_0)_L.$

Choose $N_1$ so $\sum_{k\in A_0,k\le N_1}a_k\ge1$. Let

$$A_1=\{k\in A_0:k>N_1,b_k\in(I_1)_L\},\quad B_1=\{k\in A_0:k>N_1,b_k\in(I_1)_R\}.$$

And so on. I'll let you write out the details formally, it being your problem after all. (Probably in a formal writeup we'd need a sequence $C_n$, which equuals $A_n$ or $B_n$ for every $n$.) We end up with a sequence of intervals $I_n$ with $|I_n|=2^{-n}$ and $I_{n+1}\subset I_n$ and a sequence of pairwise disjoint sets of integers $S_n$ so that if we let $S=\bigcup S_n$ then everything works:

For each $n$, $\{k\in S:b_k\notin I_n\}$ is finite. This shows that the limit of $b_k$ exists, as $k\to\infty$ in $S$.

For each $n$, $\sum_{k\in S_n}a_k\ge1$; hence $\sum_{k\in S}a_k=\infty$.

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  • $\begingroup$ The step "Choose $N_1$ so $\sum_{k \in A_0, k \leq N_1} a_k$ is infinite" does not make sense, since it sums over a finite collection of real numbers. Perhaps you mean to choose enough points in that group so their sum is at least 1? $\endgroup$ – Michael Aug 13 '15 at 17:07
  • $\begingroup$ That's exactly what I meant, thanks! $\endgroup$ – David C. Ullrich Aug 13 '15 at 17:08
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Claim: The following result is easy to prove by induction.

For each $n$ there exists a positive integer $0 \leq k \leq 2^{n}-1$ such that the set $$I_n = \left\{ j \mid \frac{k}{2^n} \leq b_ j\leq \frac{k+1}{2^n} \right\}$$ is infinite and $$\sum_{k \in I_n} a_k =\infty$$

The inductive step follows from the fact that $$I_n = \left\{ j \mid \frac{2k}{2^{n+1}} \leq b_ j\leq \frac{2k+1}{2^{n+1}} \right\} \cup \left\{ j \mid \frac{2k+1}{2^{n+1}} \leq b_ j\leq \frac{2k+2}{2^{n+1}} \right\}$$ and for at least one of the sets the corresponding $a$-sum is infinite.

Construction Now, pick $k_{1} < k_2< ...< k_{i_1}$ from $I_1$ such that $$a_{k_1}+...+a_{k_{i_1}}>1$$

Inductively construct $i_n>i_{n-1}$ and $$k_{i_{n-1}+1},..,k_{i_n} \in I_n$$ such that $$k_{i_{n-1}}<k_{i_{n-1}+1}<..<k_{i_n}$$ and $$a_{k_{i_{n-1}}}+a_{k_{i_{n-1}+1}}+...+a_{k_{i_n}} >1$$

This sequence has the two properties (convergence of $b_{k_n}$ follows from the fact that each $I_n$ contains all but finitelly many $k_n$).

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