# Square Root in an alternating power series

## Main Question or Discussion Point

I had a question similar to Σ0 (-1)^n (x)^(n/2) and attempted to solve it using the root test getting abs(√x)<1, but I've also seen some places answer it as √abs(x)<1 so am I skipping a step.

BvU
Homework Helper
2019 Award
Hello again,
You want to read your new thread as if you were an interested helper. At this moment
I had a question similar to Σ0∞ (-1)^n (x)^(n/2) and attempted to solve it using the root test getting abs(√x)<1, but I've also seen some places answer it as √abs(x)<1 so am I skipping a step.
doesn't make much sense to me and I expect it doesn't to many others either. What is this about and what is your question ?

I'm sorry about that, I was trying to find the interval of convergence of ∑ 0, n→∞ (-1)^n*(x)^(n/2), I got √x<1 or [0,1) after using the root test but when I checked Wolfram it said the absolute value of √x or (1,1) and I tried to find a similar question with the steps involved but couldn't, I hope it makes more sense now, I tried to proofread it but thats not my strong suit.

BvU
Homework Helper
2019 Award
do the terms look familiar if I rewrite as$$\sum_{{\rm n}=0}^\infty \left (-\sqrt x \right )^n \ \ \rm ?$$

do the terms look familiar if I rewrite as$$\sum_{{\rm n}=0}^\infty \left (-\sqrt x \right )^n \ \ \rm ?$$
$$\sum_{{\rm n}=0}^\infty \left (-1 \right )^n \left (\sqrt x \right )^\left (n/2 \right ) \ \ \rm ?$$
sorry, not great at putting in equations just figuring it out hope it's less vague now, also I just realized the root test would make the -1 a moot point anyway, thanks for trying to work with my vagueness, also if I haven't made it clear I'm trying to figure out if the interval of integration is 0 to 1 or -1 to 1.

BvU
Hold it, post #1 doesn't have $\displaystyle \sum_{{\rm n}=0}^\infty \left (-1 \right )^n \left (\sqrt x \right )^\left (n/2 \right ) \ \ \rm$ but (from $x^{n\over 2}$ ) $$\sum_{{\rm n}=0}^\infty \left (-1 \right )^n \left (\sqrt x \right )^n\ \ = \sum_{{\rm n}=0}^\infty \left (- \sqrt x \right )^n\ \ \rm !$$
Hold it, post #1 doesn't have $\displaystyle \sum_{{\rm n}=0}^\infty \left (-1 \right )^n \left (\sqrt x \right )^\left (n/2 \right ) \ \ \rm$ but (from $x^{n\over 2}$ ) $$\sum_{{\rm n}=0}^\infty \left (-1 \right )^n \left (\sqrt x \right )^n\ \ = \sum_{{\rm n}=0}^\infty \left (- \sqrt x \right )^n\ \ \rm !$$