I have a question about series for calculus 2...?

Question

Determine whether the series is absolutely convergent, conditionally convergent or divergent.

a) n=1 Σ ∞ (-1)^n * 2n/(4n^4 +3)^.5

b) n=1 Σ ∞ (-1)^n * (2^n)/(n!)

any help would be appreciated!

Answer 1

sorry comparison test is NOT applicable to alternate series
a) in absolute value the series is of the same class as 1/n which is divergent
so you have to proof that in abs val. it is monotone decreasing
Take the function
f(x) = x/(4x^4+3)^0.5 and calculate the derivative
f´(x) = 1/(4x^4+3) * [(4x^4+3)^0.5 -x*(16x^3)/(2*(4x^4+3)^0.5]
The sign,for positive x depends on (4x^4+3 -8x^4)=(-4x^4+3)
which is negative for x>1
so 2n/(4n^4+3)^0.5 is decreasing and by the criteria for alternate series is conditionally convergent

b) taken in absolute value an+1/an = 2/(n+1) ==>0<1 so the series is absolutely convergent

Answer 2

a) conditionally convergent
b) absolutely convergent
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Reason:
a) compare the series to the series of 1/n
b) a(n+1)/a(n) = 2/(n+1) < 1 when n is larger than 1.