# Series convergence test, $\sum_{n=1}^{\infty} \frac{(x-2)^n}{n3^n}$

I'm trying to find all $$x$$ for which $$\sum_{n=1}^{\infty} \frac{(x-2)^n}{n3^n}$$ converges. I know I need to check the ends ($$-1$$ and $$5$$) but I'm not sure what to happen after that. I'm pretty sure I'd substitute the values of $$x$$ into the sums and then I'd use convergence tests to see what works, but I always get stuck.

Apparently, I'm supposed to get the alternating harmonic series test for the $$-1$$ and the harmonic series test for $$5$$ but I'm unable to manipulate the series to get this. I've tried ratio tests but they don't simplify into what I want.

Actually, I figured it out... I was writing down $$x+2$$ rather than $$x-2$$ and now it all makes sense.

• Your problem is not very clear tome. Did you obtain the radius of convergence? Jun 28, 2020 at 10:30

By the ratio test, every x value between -1 and 5 would make the series converge.
we just need to find out whether x=-1, 5 makes it converge.

1. x=-1: The series will look like this. $$\sum_{n=1}^{\infty} (-1)^n/n$$ The series without $$(-1)^n$$ is 1. always positive, 2. the limit of it is zero, 3. and it is a decreasing sequence. Therefore, by the alternating series test, it converges.
2. x=5: The series will look like this. $$\sum_{n=1}^{\infty} 1/n$$ We know that this series diverges by the p-series test.
So, the interval of convergence would be $$-1\leq x< 5$$

It's Taylor series expansion for $$\log (\frac{1}{1-z})$$, with $$z = \frac{x-2}{3}$$. So it converges for $$|\frac{x-2}{3}|<1$$

• Pls see the edit
– Alex
Jun 28, 2020 at 13:06

Let $$a_n = \frac{(x-2)^n}{n3^n}$$, then $$\limsup {|a_n|^{\frac{1}{n}}} = \lim \left| \frac{x-2}{n^{\frac{1}{n}}3} \right| = \left|\frac{x-2}{3}\right|$$. Setting $$\left|\frac{x-2}{3}\right|<1$$, we get that the given series converges absolutely for $$|x-2| < 3$$ and diverges for $$|x-2| >3$$. Now at $$x = -1$$, the terms go to zero and the series is alternating. Therefore it converges at $$-1$$. At $$x = 5$$, it's the harmonic series which diverges, so it doesn't converge there.