3
$\begingroup$

Find the values of x for which the series

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

converge?

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $0$.

$\endgroup$

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.