1. ## alternating series test

$\sum^{\infty}_{n=3} \frac{cos(n\pi)}{n^{\frac{3}{5}}}$

What would I compare this to? $\frac{cos(n)}{n}$ ?

Also, when n=3 how does this change things?

Thanks!

2. Notice that: $\cos (n\pi) = (-1)^{n}$

So really, your series is just: $\sum_{n=3}^{\infty} \frac{(-1)^n}{n^{\frac{3}{5}}}$

And there really isn't anything special about starting from $n = 3$.

3. Originally Posted by o_O
Notice that: $\cos (n\pi) = (-1)^{n}$

So really, your series is just: $\sum_{n=3}^{\infty} \frac{(-1)^n}{n^{\frac{3}{5}}}$

And there really isn't anything special about starting from $n = 3$.

Ok, so I would say that:

$\sum a_n > 0$ so it converges and since $0 < \sum a_{n+1} < \sum a_n$, then $\sum a_{n+1}$ also converges?