Question

Question #c0aff

Answer 1

`sum_(n=0)^oo(2^nsin^n(x))/n^2` converges if and only if

`x in [kpi-pi/6, kpi+pi/6]` for some `k in ZZ`

Explanation:

First, note that `(2^nsin^n(x))/n^2 = (2sin(x))^n/n^2`.

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If `|2sin(x)| <= 1` then `|(2sin(x))^n/n^2| <= 1/n^2`.

As `sum_(n=0)^oo1/n^2` is a known convergent series, this implies `sum_(n=0)^oo(2sin(x))^n/n^2` converges by the comparison test.

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If `|2sin(x)| > 1` then `lim_(n->oo)|(2sin(x))^n/n^2|` diverges (this can be demonstrated through two applications of L'Hopital's rule ) and thus `sum_(n=0)^oo(2sin(x))^n/n^2` diverges by the divergence test.

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By the above, we can say that `sum_(n=0)^oo(2^nsin^n(x))/n^2` converges if and only if `|2sin(x)|<= 1`, that is, when `|sin(x)| <= 1/2`.

Solving for `x` gives us our final result:

`sum_(n=0)^oo(2^nsin^n(x))/n^2` converges if and only if

`x in [kpi-pi/6, kpi+pi/6]` for some `k in ZZ`