need some help with a power series convergence test problem

Find the interval of convergence for the given power series: $$\sum\limits_{n=1}^\infty \frac{(x - 1)^n }{n(-4)^n}$$

First I applied the generalized ratio test, came out with $\frac{(1-x)}{4}$

Solved the inequality $|1-x| \lt 4$ and got $-3 \lt x \lt 5$.

But webwork refuses to accept my answer. Am I doing something wrong?

• It converges when $x=5$. – Hanul Jeon May 3 '15 at 5:08

The ratio test is inconclusive when the limit of the ratio is 1. So for $-3<x<5$, the ratio is less than 1, so the series converges. Now for $-3$ and $-5$, you need to check separately. For 5, the series converges to $\ln 2$. For $-3$ it diverges, since it becomes a harmonic series. So the interval of convergence is $x \in (-3, 5]$.

• Ahhh, so close. I was about to loose it for real, thanks! – CCarlos May 3 '15 at 5:16
• @CCarlos consider accepting, so other people see this as the right answer when they visit this question. – SalmonKiller May 3 '15 at 5:21

You can apply ration test only when $\lim |a_{n+1}/a_n|$ is not equal to 1. If it is equal to 1, it does not tell whether $\sum a_n$ converges or not.

So, in that case, you should check the convergence of the series directly. If $x=-3$, then it is same to the harmonic series so diverges. If $x=5$ then it converges by alternating test.

Hint

You could make life easier writing $$S=\sum_{n=1}^\infty \frac{(x - 1)^n }{n(-4)^n}=\sum_{n=1}^\infty (-1)^n\frac{y^n}n=-\log(1+y)$$ where $y=\frac{x-1}4$ must be such that $|y|<1$.

I am sure that you can take from here.

$$\sum\limits_{n=1}^{\infty} \frac{(x - 1)^n }{n(-4)^n}$$ Using the ratio test, we have $$\lim\limits_{n\to\infty} \left|\frac{\frac{(x - 1)^{n+1}}{(n+1)(-4)^{n+1}}}{\frac{(x - 1)^n }{n(-4)^n}}\right|$$ $$=\lim\limits_{n\to\infty} \left|\frac{(x - 1)^{n+1}n(-4)^n}{(x-1)^n(n+1)(-4)^{n+1}}\right|$$ $$=\lim\limits_{n\to\infty} \left|\frac{(x - 1)n}{(n+1)(-4)}\right|$$ $$=\frac14\left|x-1\right|\lim\limits_{n\to\infty} \left|\frac{n}{n+1}\right|$$ $$=\frac14\left|x-1\right|\lim\limits_{n\to\infty} \left|\frac{1}{1+\frac1n}\right|$$ $$=\frac14\left|x-1\right|$$ Now let's find the interval of convergence $$\frac14\left|x-1\right|\lt 1$$ $$\left|x-1\right|\lt 4$$ $$-4\lt x-1\lt 4$$ $$-3\lt x\lt 5$$ At $x=-3$, we have $$\sum\limits_{n=1}^{\infty} \frac{(-4)^n }{n(-4)^n}= \sum\limits_{n=1}^{\infty} \frac{1}{n}\Rightarrow \mbox{diverges}$$ At $x=5$, we have $$\sum\limits_{n=1}^{\infty} \frac{4^n }{n(-4)^n}= \sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n}\Rightarrow \mbox{converges}$$ Therefore, the interval of convergence is $$-3\lt x\leq 5$$