I am trying to determine whether the integral is convergent or divergent.
here is what I have got so far:
1/2y > or = 1/(cos y+2y)
So, 3/2y > or = 3/(cos y +2y)
my prediction is it should be divergent as the exponent of 2y is 1 ( According to int 1to infinite 1/x^p)
However, based on the direct comparison test, it should be converges as f(x)> g(x)...
what did I miss??
here is what I have got so far:
1/2y > or = 1/(cos y+2y)
So, 3/2y > or = 3/(cos y +2y)
my prediction is it should be divergent as the exponent of 2y is 1 ( According to int 1to infinite 1/x^p)
However, based on the direct comparison test, it should be converges as f(x)> g(x)...
what did I miss??
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This diverges by the Comparison Test.
Details:
For all y > 1, we have 3/(cos y + 2y) ≥ 3/(-1 + 2y).
Since ∫(y = 1 to ∞) dy/(2y - 1) = (1/2) ln|2y - 1| {for y = 1 to ∞} = ∞,
the integral in question must also diverge.
I hope this helps!
Details:
For all y > 1, we have 3/(cos y + 2y) ≥ 3/(-1 + 2y).
Since ∫(y = 1 to ∞) dy/(2y - 1) = (1/2) ln|2y - 1| {for y = 1 to ∞} = ∞,
the integral in question must also diverge.
I hope this helps!
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g(x) is convergent if g(x) < f(x) and f(x) is convergent. You have f(x) is divergent.