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 $\displaystyle \sum (-1)^n u_n = u_1 - u_2 + u_3 - u_4 + ...$ 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?

$\displaystyle \sum_{n=1}^{\infty} \frac{\cos n}{n^2}$.

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?

Re: Can you use the Alternating Series Test on series that changes signs every 2 term

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Originally Posted by

**mathDad** The Alternating Series Test (AST) says than an alternating series $\displaystyle \sum (-1)^n u_n = u_1 - u_2 + u_3 - u_4 + ...$ 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?

$\displaystyle \sum_{n=1}^{\infty} \frac{\cos n}{n^2}$.

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.

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: $\displaystyle \sum_{n=1}^{\infty} \frac{\cos n}{n}$.

Re: Can you use the Alternating Series Test on series that changes signs every 2 term

Quote:

Originally Posted by

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

You mean $\displaystyle{\sum_{n=1}^\infty} \dfrac {\cos(n)} n ?$

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.)

Re: Can you use the Alternating Series Test on series that changes signs every 2 term

Define a sequence:

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

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

$\displaystyle k \in \left[1+\left\lfloor \dfrac{3\pi}{2} + 2(n-1)\pi \right\rfloor, \left\lfloor \dfrac{\pi}{2} + 2n\pi \right\rfloor \right]$, $\displaystyle \cos k > 0$.

Hence, $\displaystyle \sum_{n=1}^\infty \dfrac{\cos n}{n} = \sum_{n=0}^\infty a_n = \sum_{n=0}^\infty (-1)^n |a_n|$. So, you can use the alternating series test on that. You just need to show that $\displaystyle \lim_{n \to \infty} |a_n| = 0$.

Re: Can you use the Alternating Series Test on series that changes signs every 2 term

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Originally Posted by

**romsek** You mean $\displaystyle{\sum_{n=1}^\infty} \dfrac {\cos(n)} n ?$

Re-add the post you just deleted. It had a useful link and you directly answered the question.

Re: Can you use the Alternating Series Test on series that changes signs every 2 term

Oops, I messed up $\displaystyle a_n$ slightly. It should actually be:

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