# Can you use the Alternating Series Test on series that changes signs every 2 terms?

• April 20th 2014, 09:26 AM
Can you use the Alternating Series Test on series that changes signs every 2 terms?
The Alternating Series Test (AST) says than an alternating series $\sum (-1)^n u_n = u_1 - u_2 + u_3 - u_4 + ...$ converges if the [tex]u_n[\tex]'s are all positive, decreasing and its limit is 0. Does that exclude using it on terms that alternate every two or three terms? Even if it switches signs reliably?

Does this series converge or diverge?
$\sum_{n=1}^{\infty} \frac{\cos n}{n^2}$.

This series alternates sign every third term, but all the other conditions are met. If it doesn't satisfy the AST, then how would you determine it's convergence?
• April 20th 2014, 09:49 AM
romsek
Re: Can you use the Alternating Series Test on series that changes signs every 2 term
Quote:

The Alternating Series Test (AST) says than an alternating series $\sum (-1)^n u_n = u_1 - u_2 + u_3 - u_4 + ...$ converges if the [tex]u_n[\tex]'s are all positive, decreasing and its limit is 0. Does that exclude using it on terms that alternate every two or three terms? Even if it switches signs reliably?

Does this series converge or diverge?
$\sum_{n=1}^{\infty} \frac{\cos n}{n^2}$.

This series alternates sign every third term, but all the other conditions are met. If it doesn't satisfy the AST, then how would you determine it's convergence?

this isn't a good choice for an example since $\sum \dfrac 1 {n^2}$ converges

$\left|\cos(x)\right| \leq 1$ so clearly $\sum \dfrac {\cos(x)}{n^2}$ converges

You don't have to invoke the fact that it's an alternating series to get convergence. This series converges absolutely.
• April 20th 2014, 09:54 AM
Re: Can you use the Alternating Series Test on series that changes signs every 2 term
OK. How about the same series with n on the bottom: $\sum_{n=1}^{\infty} \frac{\cos n}{n}$.
• April 20th 2014, 09:57 AM
romsek
Re: Can you use the Alternating Series Test on series that changes signs every 2 term
Quote:

OK. How about the same series with n on the bottom: $\sum_{n=1}^{\infty} \frac{\cos n}{n^2}$.

You mean $\displaystyle{\sum_{n=1}^\infty} \dfrac {\cos(n)} n ?$
• April 20th 2014, 10:00 AM
Re: Can you use the Alternating Series Test on series that changes signs every 2 term
Yes. I posted too quick :(. (Will correct my post.)
• April 20th 2014, 10:34 AM
SlipEternal
Re: Can you use the Alternating Series Test on series that changes signs every 2 term
Define a sequence:

$a_n = \begin{cases}\cos 1 & \text{if }n=0 \\ \displaystyle \sum_{k=1+\left\lfloor \tfrac{\pi}{2} + 2(n-1)\pi \right\rfloor}^{\left\lfloor \tfrac{3\pi}{2} + 2(n-1)\pi \right\rfloor} \dfrac{\cos k}{k} & n \equiv 1 \pmod{2} \\ \displaystyle \sum_{k=1+\left\lfloor \tfrac{3\pi}{2} + 2(n-1)\pi \right\rfloor}^{\lfloor \tfrac{\pi}{2} + 2n\pi \right\rfloor} \dfrac{\cos k}{k} & \text{otherwise}\end{cases}$

Note that for each $k \in \left[1+\left\lfloor \dfrac{\pi}{2} + 2(n-1)\pi\right\rfloor,\left\lfloor \dfrac{3\pi}{2} + 2(n-1)\pi\right\rfloor\right]$, $\cos k < 0$ and for each
$k \in \left[1+\left\lfloor \dfrac{3\pi}{2} + 2(n-1)\pi \right\rfloor, \left\lfloor \dfrac{\pi}{2} + 2n\pi \right\rfloor \right]$, $\cos k > 0$.

Hence, $\sum_{n=1}^\infty \dfrac{\cos n}{n} = \sum_{n=0}^\infty a_n = \sum_{n=0}^\infty (-1)^n |a_n|$. So, you can use the alternating series test on that. You just need to show that $\lim_{n \to \infty} |a_n| = 0$.
• April 20th 2014, 10:45 AM
You mean $\displaystyle{\sum_{n=1}^\infty} \dfrac {\cos(n)} n ?$
Oops, I messed up $a_n$ slightly. It should actually be:
$a_n = \begin{cases}\cos 1 & \text{if }n=0 \\ \displaystyle \sum_{k=1+\left\lfloor n\pi - \tfrac{\pi}{2}\right\rfloor}^{\left\lfloor (n+1)\pi -\tfrac{\pi}{2} \right\rfloor} \dfrac{\cos n}{n} & \text{otherwise}\end{cases}$