# Where am I making my mistake? (intervals of convergence)

The sum is: \begin{align} \sum_{n=1}^{\infty} \frac{(x-6)^n}{(-8)^nn} \end{align} I end up with $|(x - 6)/8| < 1$ and therefore, $-8 < 6 - x < 8$ so $14 > x > -2$, but that gives me the wrong answer.
I've been up all night doing this homework so chances are I'm missing something obvious or committing a rookie mistake. Thank you for your help.

-
I changed the sum to latex, please check if it is still correct. And maybe start to formulate the question. –  sonystarmap Dec 17 '12 at 13:45
@macydanim THank you very much! –  Ceelos Dec 17 '12 at 13:48

## 1 Answer

You have derived $-8<6-x<8$. Subtracting $6$ from each term gives $-14<-x<2$. Now negating all the terms gives $-2<x<14$.

So now we check if the sum converges at the boundary points of the interval. When $x = 14$, the sum is $\sum \frac{(-1)^n}{n}$, which converges. When $x = -2$, the sum is $\sum \frac{1}{n}$, which diverges. The sum should then converge on $-2<x\leq 14$.

-
yes, sorry, that was a mistake. I meant to write -2. –  Ceelos Dec 17 '12 at 13:48
My homework website must've been acting up (it'll do this sometimes) so it must've not been accepting the correct answer. I just typed in -2 & 14 and it worked. Thank you for answering froggie. –  Ceelos Dec 17 '12 at 13:53