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What do I do when I end up with x as a square in an inequality? e.g. $-1 < (x+1)^2 < 1$ or $ -1 < 2(x+2)^2 <1$ ? Should there even be a negative interval since x is squared? How would we find the interval and radius of convergence when we end up with squares?

Edit: the original power series questions are $$\sum_{n=1}^{\infty} (-1)^n \frac{(x+1)^{2n+1}}{n^2 +4} $$ $$\sum_{n=1}^{\infty} 2^n (x+2)^{2n+1} $$

OoRoOrOoRoO
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  • You are right. For real $x$, we have $-1 < (x+1)^2 < 1$ if and only if $0 \le (x+1)^2 < 1$. "Radius of convergence" makes sense for power series; are you doing this computation for a power series, or not? – GEdgar Mar 23 '20 at 19:35
  • @GEdgar yes, it's for a power series. i posted the original two questions above – OoRoOrOoRoO Mar 23 '20 at 20:45

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It can shown as $|(x+1)|^2 < 1$ and the second one can be shown as $ 2|(x+2)|^2 < 1$. From there, we can solve and the two of them become $ -2 < x < 0 $ and $ - \frac{1}{\sqrt{2}} -2 < x < \frac{1}{\sqrt{2}} -2 $ respectively.

OoRoOrOoRoO
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