Square Root in an alternating power series

  • #1

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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.
 

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  • #2
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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 ? :smile:
 
  • #3
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.
 
  • #4
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do the terms look familiar if I rewrite as$$
\sum_{{\rm n}=0}^\infty \left (-\sqrt x \right )^n \ \ \rm ?$$
 
  • #5
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.
 
  • #6
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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}## ) $$
\sum_{{\rm n}=0}^\infty \left (-1 \right )^n \left (\sqrt x \right )^n\ \ = \sum_{{\rm n}=0}^\infty \left (- \sqrt x \right )^n\ \ \rm !
$$
 
  • #7
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}## ) $$
\sum_{{\rm n}=0}^\infty \left (-1 \right )^n \left (\sqrt x \right )^n\ \ = \sum_{{\rm n}=0}^\infty \left (- \sqrt x \right )^n\ \ \rm !
$$
I didn't notice the square root and copied yours to figure out the equation text so didn't notice there either really sorry about making this more complicated than it should be, used to using fractional exponents in sums not roots, looks like you had it right.
 

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