# Alternating series remainder proof

For an alternating series $\sum_{n=1}^{\infty}(-1)^n a_n$, where $a_n \geq 0$ is decreasing to $0$, we know $$S_{2n-1} \leq S \leq S_{2n}$$

where $S_n$ are the partial sums, then $$|S- S_n| \leq |S_{n+1} - S_n| = a_{n+1}$$

My question is how exactly do we get from the first (double) inequality to the second?

which shows the first five values of $s_n$, and imagine that $S$ is between $S_3$ and $S_4$. You can see that the formula $$|S-S_n|\le|S_{n+1}-S_n|$$ is true because on the left side you have the distance between $S$ and $S_n$, which is less than (or equal to) the distance between $S_{n+1}$ and $S_n$. This claim descends from the monotonicity of $S_n$.
And $|S_{n+1}-S_n|$ is equal to $a_{n+1}$ for the definition of $S_n$.
• +$1$: This is exactly how I describe it to my students. – Clayton Jun 29 '18 at 15:32