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_{0}

^{∞}(-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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- Thread starter Satirical T-rex
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BvU

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You want to read your new thread as if you were an interested helper. At this moment

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 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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BvU

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\sum_{{\rm n}=0}^\infty \left (-\sqrt x \right )^n \ \ \rm ?$$

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$$

\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.

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\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

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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.

\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 !

$$

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