Absolute Convergence

Question

Show whether the series ((-1)^n)*((2^n)/(n^4))) is absolutely convergent.

Taking an=((-1)^n)*((2^n)/(n^4))) , here |an|=((2^n)/(n^4)).

I have the gut feeling that this is absolutely convergent. But don't see a way to do that. Also, we can only use comparison test, monotone convegence theorem, convergent==>boundedness, convergent sequence has only one limit.

To show |an| is absolutely convergent(which will also prove an is convergent), I need to find another bn, which converges, such that an<=bn. I know bn=n!/n^4 will do, but need to prove that converges first. How can I prove this?

Maybe we can use comparison test seperately more than once right?

Any thoughts on solving this?

Thanks!

Answer 1

In order for the series ∑ (-1)ⁿ 2ⁿ / n⁴ to converge, it must hold that the sequence |(-1)ⁿ 2ⁿ / n⁴| = 2ⁿ / n⁴ converges to 0 as n → ∞.

If 2ⁿ / n⁴ does not converge to 0, then the series ∑ (-1)ⁿ 2ⁿ / n⁴ diverges.

By applying L'Hospital's Rule four times, we have that

lim_{n →∞} 2ⁿ / (n⁴)

= lim_{n →∞} ln(2) 2ⁿ / (4n³)

= lim_{n →∞} ln(2)² 2ⁿ / (12n²)

= lim_{n →∞} ln(2)³ 2ⁿ / (24n¹)

= lim_{n →∞} ln(2)⁴ 2ⁿ / 24 = ∞

Since the sequence |(-1)ⁿ 2ⁿ / n⁴| does not converge to 0, the series ∑(-1)ⁿ 2ⁿ / n⁴ diverges.

Answer 2

Why do you need to use comparison? Maybe other tests work better.

Anyway, series ∑2ⁿ/n⁴ . does not converge so the original series is not absolute convergent but we are still not sure about the the original series ∑(-1)ⁿ2ⁿ/n⁴ . It might converge due the change of sign.

But it seems that this series does not converge either