# About complex sum and modulus

Let $\left(a_{n}\right)_{n},\,\left(b_{n}\right)_{n}$ two succession of non negative real numbers, $\left(c_{n}\right)_{n}$ a succession of complex numbers and $N$ a large natural number. Suppose that $$a_{n}\leq b_{n},\,\forall n.$$ My question is: is it true that $$\left|\underset{n=1}{\overset{N}{\sum}}a_{n}c_{n}\right|\leq\left|\underset{n=1}{\overset{N}{\sum}}b_{n}c_{n}\right|?$$

## 2 Answers

Let

\begin{align*} \{a_n\}&=\{1,0,0,0,\dotsc\} \\ \{b_n\}&=\{1,1,0,0,\dotsc\} \\ \{c_n\}&=\{1,-1,0,0,\dotsc\} \end{align*} Then for $N\geq 2$ $$\left|\sum_{n=1}^Na_n\cdot c_n\right|=1$$ but $$\left|\sum_{n=1}^Nb_n\cdot c_n\right|=0$$

One could also concoct a slightly more interesting counterexample. Let \begin{align*} a_n&= \begin{cases} 0 & 1\leq n<M \\ 1 & n\geq M \end{cases} \\ b_n&=1 \\ \{c_n\}&=\{\zeta_1,\zeta_2,\dotsc,\zeta_M,0,0,\dotsc,\} \end{align*} where $\zeta_1,\dotsc,\zeta_M$ are the $M$th roots of unity. Then for $N\geq M$ $$\left|\sum_{n=1}^Na_n\cdot c_n\right|=1$$ but $$\left|\sum_{n=1}^Nb_n\cdot c_n\right|=0$$

• Thank you! And if $\left(a_{n}\right)_{n},\,\left(b_{n}\right)_{n}$ are two monotone increasing successions? – user118929 Feb 23 '14 at 20:55

Consider $(a_n) = (c_n) = -1,1,-1,1...$ and $(b_n) = 2,2,2,2,2,2...$

• $a_n$ and $b_n$ are given as nonnegative. – Berci Feb 23 '14 at 20:49
• Ah, didn't see that – Steven Gubkin Feb 23 '14 at 20:50