# Splitting a sequence

Prove the following statement.

For every sequence $x_1, x_2, \dots$ of nonnegative real numbers there exist two sequences $a_1, a_2, \ldots$ and $b_1, b_2, \ldots$ of nonnegative real numbers such that:

• $x_n=a_n+b_n$ for all $n$
• $a_1+\ldots+a_k\le k$ for infinitely many $k$
• $b_1+\ldots+b_h\le h$ for infinitely many $h$.
• It isn't true for $x_{n} = 2^n$. – gukoff Apr 23 '13 at 16:43
• @Harold: yes it is. I fooled myself for a while, too. – Ross Millikan Apr 23 '13 at 16:49
• Oh, yes. The problem is actually extremely easy, thanks. – gukoff Apr 23 '13 at 17:01

Hint: if $a_1=0$ then you get one $k$ that works. Then you must have $b_1=x_n$. What is the first $h$ that may work? Then what is the next possibility for $k$ that works?
• 1) $a_1=0$ and $b_1=x_1$ 2) $a_i=x_i$ and $b_i=0$ with $i\in\{2,\ldots, x_1\}$ 3) $a_i=0$ and $b_i=x_i$ with $i\in\{x_1+1, \ldots, x_2+\ldots +x_{x_1}\}$ and so on... Right? – user72870 Apr 23 '13 at 17:36