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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.

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

3 Answers 3

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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 $$-1\leq x< 5$$
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It's Taylor series expansion for $\log (\frac{1}{1-z})$, with $z = \frac{x-2}{3}$. So it converges for $|\frac{x-2}{3}|<1$

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  • $\begingroup$ Pls see the edit $\endgroup$
    – Alex
    Jun 28, 2020 at 13:06
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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.

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