Find out the convergence space of the following series $$\sum_{n=1}^\infty \frac{n(x-1)^n}{3^ n (2n-1)}$$ I have found that the convergece radius is equal to $R = 3 $ but i am not sure how to check if it is convergent the closed space $[-3,3]$ or the open one
Find the convergence space of $\sum_{n=1}^\infty \frac{n(x-1)^n}{3^ n (2n-1)}$
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convergence
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0Do you know how to use Ratio Test? – 2017-02-02
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0Be careful. The radius is 3, but the center is not 0. – 2017-02-02
2 Answers
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To test along the boundary, plug in the values for $x$ and see what happens. I'll do the positive one. For the negative one, the ideas are similar, although often times the alternating series test makes conditional convergence work out here.
$$\sum \limits_{n=1}^{\infty} \frac{2^n n}{3^n(2n-1)} = \sum \limits_{n=0}^{\infty} \left(\frac{2}{3}\right)^n \frac{n}{2n-1}$$
The right factor is always less than $1$, so this series is bounded above by a geometric series with sufficiently small radius, and so it converges at $x=3$. Now you do the same for $-3$, bearing in mind the comment about alternating series.
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Well, the series is a power series centred at $x=1$, so the question should be if the interval of convergence is $[-2,4]$ or $(-2,4)$ (or one of the two possible half-open intervals that would also be possible, to be exact), not whether it's $[-3,3]$ or $(-3,3)$.
The easiest way to determine this would be to just plug in the two edges of the interval into your series.
For $x=4$, we have
\begin{align} \sum_{n=1}^\infty\frac{n\cdot3^n}{3^n(2n-1)}&=\sum_{n=1}^\infty\frac{n}{2n-1} \end{align}
which diverges, as the general term doesn't go to $0$.
By a similar calculation, we have for $x=-2$:
\begin{align} \sum_{n=1}^\infty\frac{n\cdot(-3)^n}{3^n(2n-1)}=\sum_{n=1}^\infty(-1)^n\frac{n}{2n-1} \end{align}
Once again, the general term doesn't go to $0$, so your series diverges at $x=-2$ as well.
Thus the interval of convergence is $(-2,4)$.