1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ It converges when $x=5$. $\endgroup$ – Hanul Jeon May 3 '15 at 5:08
1
$\begingroup$

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 $x \in (-3, 5]$.

$\endgroup$
  • $\begingroup$ Ahhh, so close. I was about to loose it for real, thanks! $\endgroup$ – CCarlos May 3 '15 at 5:16
  • $\begingroup$ @CCarlos consider accepting, so other people see this as the right answer when they visit this question. $\endgroup$ – SalmonKiller May 3 '15 at 5:21
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

$$\sum\limits_{n=1}^{\infty} \frac{(x - 1)^n }{n(-4)^n}$$ 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 $$ -3\lt x\leq 5 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.