Find the values of x for which the series

$$\sum_{n=0}^{\infty} \frac{(x-1)^n}{(-3)^n}$$


Not really sure how to properly answer this question considering its edge terms. Here goes my attempt: $$\sum_{n=0}^{\infty} \left(\frac{x-1}{-3}\right)^n$$

I know by the geometric series test that the only way for this geometric serise to be convergent is if $|r| < 1$

$$|x-1| < 3$$

$$\leftrightarrow -3 < (x-1) < 3$$

$$\leftrightarrow -2 < x < 4$$

Therefore this series converges for $x \in (-2,4)$

Is this a good enough answer? I know it doesn't converge for 4 or -2.

  • $\begingroup$ typo on your final statement, you wrote $x-1$, think you meant $x$. And simply, yes it is a good enough answer for me at least. $\endgroup$ – Robin Aldabanx Mar 29 '17 at 3:35
  • $\begingroup$ You can use ratio test ? $\endgroup$ – BAYMAX Mar 29 '17 at 3:36
  • $\begingroup$ Never learned ratio test yet. And thank you robin $\endgroup$ – user349557 Mar 29 '17 at 3:36
  • $\begingroup$ Yeah it looks good! $\endgroup$ – BAYMAX Mar 29 '17 at 3:38

You close by saying you don't know if it converges when $x = 4$ or $x = -2$.

At these values, note that $x-1 = 3$ or $-3$, respectively, so that in the former case you would be considering the infinite series $1-1+1-1+\cdots$ and in the latter you have $1+1+1+\cdots$

Each of these series diverges, as the $n^{\text{th}}$ terms do not converge to $0$.


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