2
$\begingroup$

enter image description here

Hi! I am currently working on some calc2 online homework problems on Taylor series and Maclaurin series. I have tried a few different answers to this question, but I am really not sure how to go about solving this problem. If anyone can help me I would greatly appreciate it!

$\endgroup$
1
$\begingroup$

Observe that $$\frac{1}{1 - 3x} = \frac{1}{-2 - 3(x - 1)} = -\frac{1}{2 + 3(x - 1)} = -\frac{1/2}{1 + (3/2)(x - 1)}.$$ Therefore $$\frac{1}{1 - 3x} = -\frac{1}{2}\sum_{n = 0}^\infty (-1)^n \left(\frac{3(x - 1)}{2}\right)^n = -\frac{1}{2}\sum_{n = 0}^\infty (-1)^n \left(\frac{3}{2}\right)^n (x - 1)^n, \quad \left|\frac{3(x - 1)}{2}\right| < 1.$$

Because $$\left|\frac{3(x - 1)}{2}\right| < 1 \Longleftrightarrow \frac{1}{3} < x < \frac{5}{3},$$ and the series above diverges when $x = 1/3$ or $x = 5/3$, we conclude that the interval of convergence is $(1/3, 5/3)$.

$\endgroup$

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.