Suppose that [tex]\sum[/tex]anxn has finite radius of convergence R and that an >= 0 for all n. Show that if the series converges at R, then it also converges at -R.
The Attempt at a Solution
Since the series converges at R, then I know that [tex]\sum[/tex]anRn = M.
At -R, the series is the following: [tex]\sum[/tex]an(-R)n = [tex]\sum[/tex](-1)nanRn.
I'm not sure where to go from here. I thought I needed to use the alternating series test, but how can I know that a1 >= a2 >= ... >= an for all n? Do I know this because the series converges? Thanks for your help.