0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

Have a look at this picture:

enter image description here

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 $S_n$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ +$1$: This is exactly how I describe it to my students. $\endgroup$ – Clayton Jun 29 '18 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.