Absolute value of infinite series sum

Question

How does it come about that $$\left|\Sigma_{n=-N}^{N}c_n(f)e^{inx} - \Sigma_{-\infty}^{+\infty} c_n(f)e^{inx}\right| = \left|\Sigma_{|n|>N} c_n(f)e^{inx}\right|?$$ What happens with the $n$-index? Could someone please show steps involved?

Answer 1

\begin{align*} \left|\Sigma_{n=-N}^{N}c_n(f)e^{inx} - \Sigma_{-\infty}^{+\infty} c_n(f)e^{inx}\right| &= \left|-\Sigma_{-\infty}^{-(N+1)} c_n(f)e^{inx}-\Sigma_{N+1}^{+\infty} c_n(f)e^{inx}\right|\\ &=\left|-\left(\Sigma_{-\infty}^{-(N+1)} c_n(f)e^{inx}+\Sigma_{N+1}^{+\infty} c_n(f)e^{inx}\right)\right|\\ &=\left|\Sigma_{|n|>N} c_n(f)e^{inx}\right|. \end{align*}

As you can see, you are essentially removing the 'center' (around $0$) of the infinite summation. You then combine both terms by noting that $|-n|=|n|>N$ by using the clever notation $|n|>N$.