The first part of the proof of the error estimate theorem in integral calculus is confusing me. It states that $$\biggr\vert \sum_{n=0}^{\infty}(-1)^nb_n-\sum_{n=0}^N(-1)^nb_n\biggr\vert=\biggr\vert \sum_{n=N+1}^{\infty}(-1)^nb_n)\biggr\vert$$
I don't understand why the lower bounds reduce to $N+1$. I've tried drawing a graph where the x-axis represents the bounds of summation and the y-axis represents the sum. My domain starts at $x=0$, then to the right extends to $x=N$, and then it continues for $x>N$. I drew two identical decreasing curves $s$, existing on $[0, \infty)$, and $s_N$, existing on $[0, N]$. When I tried $s-s_N$ I geometrically got a lower bound of $n=N$ for my summation rather than $n=N+1$
Note that$$\sum_{n=0}^\infty(-1)^nb_n=b_0-b_1+b_2-b_3+\cdots+(-1)^{N-1}b_{N-1}+(-1)^Nb_N+(-1)^{N+1}b_{N+1}+\cdots,$$whereas$$\sum_{n=0}^N(-1)^nb_n=b_0-b_1+b_2-b_3+\cdots+(-1)^{N-1}b_{N-1}+(-1)^Nb_N.$$So, if you subtract that second sum from the first one, every term of the form $(-1)^kb_k$ with $k\leqslant N$ will disappear and what remains is$$(-1)^{N+1}b_{N+1}+(-1)^{N+2}b_{N+2}+\cdots=\sum_{n=N+1}^\infty(-1)^nb_n.$$