Series of 2^(1/n)-1 with limit comparison test, calculus 2 tutorial

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Does this Series Converge or Diverge? Series (2^(1/n)-1)

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I was trying to figure out a nice way to explain this problem to my calculus II class, since I don't like the idea of just guessing the right comparison. It turns out that you can gain intuition for why you should compare to the harmonic series by first using the power series expansion for e^x. Using this, we have that 2^(1/n) = e^(1/n ln(2)) ≈ 1 + 1/n ln(2). Therefore 2^(1/n) - 1 ≈ 1/n ln(2) for large n. We could also use the linearization of 2^x at x=1 to get an idea of how 2^(1/n) behaves.

dirkdugan

If the series contains a composite function (e.g. 2^(1/x) as in this video), is one of its inner functions (1/x here) always a valid argument for the comparison test?

josephlauletta

u can just use integral test. f(x)=2^(1/x) - 1 is monotone decreasing for x>1, and integral 1 to infinity of 1 diverges, so the series diverges. lmk if this is flawed but that's the first way to do it that i saw

YorangeJuice

OMG He has a beard and doesn't use pens
I'm shaken ;)

_franci

we do not know lopital yet so we cannot use it

aashsyed

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