For what values of k does the series ∑n=1 to infinity (11/(2k^2 + 3))^n converge?

I think this is k < -2 and k > 2.

1 answer

this is just a geometric series, with

r = 11/(2k^2+1)

so, it converges when r < 1 or
2k^2+1 > 11
k^2 > 5
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