# Show that if $A$ and $B$ are both summable then $\sqrt{A^2 + B^2}$ is summable.

\textbf{Problem 6} Suppose that $A$ and $B$ are non-negative sequences.

1. Show that if $A$ and $B$ are both summable then $\sqrt{A^2 + B^2}$ is summable.

The below are what I have done so far. I am very unsure of the correctness of the following work. I appreciate any advice or corrections!

PRoof: Suppsoe $A$ and $B$ are non-negative sequences and both are summble. By Problem 5, we know that $A^2$ and $B^2$ are summable. In other words, $\sum A^2$ and $\sum B^2$ converges. Hence, $\sum (A^2+B^2)$ is an increasing seqeunce and it converges to some $L\in \mathbb{R}.$ It follows that $\sum (A^2+B^2)$ is bounded above by some $K$ where $K>1.$ Since $A$ and $B$ are non-negative sequences, $\sqrt{A^2 + B^2}\geq 0$ and thus the partial sum seqeunce of $\sqrt{A^2 + B^2}$ is increasing. By Monotone Convergence Theorem, if we can prove the partial sum seqeunce of $\sqrt{A^2 + B^2}$ is bouded, we are done. Let m be an arbitrary real number. Note that \

\begin{align*}\sqrt{A(m)^2 + B(m)^2}&=\sqrt{a_1^2 + b_1^2}+\sqrt{a_2^2 + b_2^2}+\cdots +\sqrt{a_m^2 + b_m^2}\\ &\leq \sqrt{(a_1^2 + b_1^2+a_2^2 + b_2^2+\cdots + a_m^2 + b_m^2)\cdot m}\\&<\sqrt{K\cdot m} \end{align*} Hence, we conclude that $\sqrt{A^2 + B^2}$ is bouded as desired.

• I am afraid that does not work. Your last step does not show that the sequence $\sqrt{ A_m^2+B_m^2}$ is bounded, as the right hand side grows arbitrarily big as $m\to \infty$. HINT: prove that for all $a, b\ge 0$, one has $\sqrt{a^2+b^2}\le a+b$. This can be interpreted geometrically via Pythagoras's theorem. Use this inequality to prove the theorem. – Giuseppe Negro Oct 9 '14 at 23:09
• Thank you. I just figured it out under the hint you point out! – Kun Oct 9 '14 at 23:38