# About complex sum

Let $\left(c_{n}\right)_{n},\,\left(d_{n}\right)_{n}$ two successions of complex numbers and let $N$ a large natural number.Is it true that $$\left|\underset{n=1}{\overset{N}{\sum}}c_{n}d_{n}\right|\leq\underset{n=1,\dots,N}{\max}\left|c_{n}\right|\left|\underset{n=1}{\overset{N}{\sum}}d_{n}\right|?$$It would be fine $$\left|\underset{n=1}{\overset{N}{\sum}}c_{n}d_{n}\right|=O\left(\underset{n=1,\dots,N}{\max}\left|c_{n}\right|\left|\underset{n=1}{\overset{N}{\sum}}d_{n}\right|\right).$$ It is important to me that the sum remains in the modulus. Thank you.

• What are your thoughts? What have you tried? – Calvin Lin Mar 3 '14 at 21:39
• @CalvinLin I tried to use Holder inequality but I'm not able to keep the sum in the modulus. – user118929 Mar 3 '14 at 21:42
• $c_n = d_n = (-1)^n$ explains why. – Daniel Fischer Mar 3 '14 at 22:13

## 1 Answer

The answer is no. If the $d_n$ are such that $d_n\ne0$, $1\le n\le N$, and $\sum_{n=1}^Nd_n=0$, it would imply that $$\sum_{n=1}^nc_n\,d_n=0$$ for all possible choices of $c_n$, which is clearly imposible (choose $c_n=\overline{ d_n}$.)