If the power series Σc_n(x^n) diverges for x=2, then it also diverges for x=3 and for x= -3?
Is the statement True or False? If true explain why. If false give counter example.
Is the statement True or False? If true explain why. If false give counter example.
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It's true. The power series is centered at 0, so if f(2) diverges then the radius of convergence must be less than or equal to 2. The series can only converge with |x| <= r <= 2, and this test fails for both x=3 and x=-3.
Edit: There is nothing in the problem that allows deducing that r=1, by the way. Let c_n = e^(n!) and the series indeed diverges at x=2...and at every other nonzero value of x. The radius of convergence of that series is r=0.
Edit: There is nothing in the problem that allows deducing that r=1, by the way. Let c_n = e^(n!) and the series indeed diverges at x=2...and at every other nonzero value of x. The radius of convergence of that series is r=0.
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If -1 < X < 1, then it will converge
If you use the root test, you get c^(1/n) * n^(1/n) * X.
As n approaches infinity c^(1/n) and n^(1/n) will both approach 1, so it will approach X. Therefore, if X < 1, it will converge, if X > 1 it will diverge, and if X = 1 it is inconclusive.
If you use the root test, you get c^(1/n) * n^(1/n) * X.
As n approaches infinity c^(1/n) and n^(1/n) will both approach 1, so it will approach X. Therefore, if X < 1, it will converge, if X > 1 it will diverge, and if X = 1 it is inconclusive.