# Interval of Converge for a Power Series

I would really appreciate some help on this problem that I have been working on.
It's a power series question. I have to find the interval of convergence for this power series.

The power series is:

$$\sum_{n=0}^\infty n(x-9)^n$$

The work I have done so far is:

I used the Ratio Test to find the radius of convergence

$\lim_{n\to \infty} |{a_{n+1}\over a_{n}}| = \lim_{n\to \infty}|{(n+1)(x-9)^{n+1}\over n(x-9)^n}| = \lim_{n\to \infty} |(x-9){n+1\over n} = lim_{n\to \infty} |x-9| = |x-9|$

According to the Ratio Test, If $|x-9| \lt 1,$ then the series will converge

I deduced that the radius, R = 9, within 1 number at each side. So the boundaries are x = 8 and x = 10. I know that at the radius the series in convergent. Now I have to test the boundaries, to check whether the series converges or diverges at those particular points.

This is the part I am having trouble with. I can't figure out if the series converges or diverges at x = 8 and x = 10. Help would be really appreciated.

At x = 8, $\sum_{n=0}^\infty n(x-9)^n$ = ?

At x = 10,$\sum_{n=0}^\infty n(x-9)^n$ = ?

• Do the terms of the series go to zero on the boundaries? – robjohn Apr 1 '14 at 4:16
• At $x=8$, $\sum_{n=0}^\infty n(x-9)^n = \sum_{n=0}^\infty (-1)^n n = 0 -1 + 2 - 3 + 4 - 5 + \cdots$. – Stephen Montgomery-Smith Apr 1 '14 at 4:30

You have convergence for $|x-9|<1 \Rightarrow -1 < x-9 < 1 \Rightarrow 8<x<10$.
One can show $\sum_{n=0}^{\infty}nx^n$ only converges for $|x|<1$.
Now your done since $|10-9| = 1, |8-9|= 1$.
$\bf{Additional}$: One can check divergence by $n^{th}$ term test.
First of all, notice that $$\sum_{n=0}^\infty n(x-9)^n=\sum_{n=1}^\infty n(x-9)^n=(x-9)\sum_{n=1}^\infty n(x-9)^{n-1}$$ The last sum is the derivative, with respect to $x$ of $$\sum_{n=1}^\infty (x-9)^{n}$$ which is geometric progression. Then $$\sum_{n=1}^\infty (x-9)^{n}=-\frac{x-9}{x-10}$$ and then $$\sum_{n=0}^\infty n(x-9)^n=\frac{x-9}{(x-10)^2}$$