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Radius convergence of 2^n * x^(2n)?
what is the radius of covergence for sum (n=0 to infinity) 2^n * x^(2n)
I figured this out to be sqrt(1/2), but how do I test the endpoints for convergence.
1 Answer
- kbLv 71 decade agoFavorite Answer
For x = sqrt(1/2), the series reduces to Σ(n=0 to ∞) 1^n, which clearly diverges.
Similarly, x = -sqrt(1/2), the series reduces to Σ(n=0 to ∞) (-1)^n, which also diverges.
So, the interval of convergence is (-sqrt(1/2), sqrt(1/2)), with radius of convergence sqrt(1/2).
I hope this helps!
