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MrMaterial

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## Homework Statement

Part a.) For a>0 Determine Lim

_{n→∞}(a

^{1/n}-1)

Part b.) Now assume a>1

Establish that Σ

_{n=1}

^{∞}(a

^{1/n}-1) converges if and only if Σ

_{n=1}

^{∞}(e

^{1/n}-1) converges.

Part c.) Determine by means of the integral test whether Σ

_{n=1}

^{∞}(e

^{1/n}-1) converges

## Homework Equations

Integral Test

Limit Comparison Test

L'hopital's rule[/B]

## The Attempt at a Solution

**Part a.)**

Looking at the limit: Lim

_{n→∞}(a

^{1/n}-1) I see that a

^{1/n}= a

^{0}as n → ∞

Lim

_{n→∞}(a

^{1/n}-1) = Lim

_{n→∞}(1-1) = 0

**To see if they both converge, I will use the limit comparison test and see if I get a value above zero but less than infinity**

**Part b.)**Lim

_{n→∞}(a

^{1/n}-1)/(e

^{1/n}-1) = Lim

_{n→∞}(1-1)/(1-1) = undefined.

Perhaps i can change the formulas to obtain a conclusive test.

I'm not exactly sure if there is anything that can be done to (a

^{1/n}-1) to make anything workable, maybe something can be done with (e

^{1/n}-1)

Notes from my instructor say that i might have to use L'hopital's rule to solve, so i will see where that takes me.

When i derive both the top and bottom I get (a

^{1/x}-1)/(e

^{1/n}-1) = (x

^{2}a

^{(1/x)}ln(a))/(x

^{2}e

^{(1/x)}) = ln(a) Lim

_{x→∞}a

^{1/x}/e

^{1/x}= 1/1 ln(a) = ln(a)

Given that ln(a) is greater than zero and less than infinity (if a>1 and a<∞) I suppose it shows that either both functions diverge or converge.

**I do not like the integral test but it's time to face my fears!**

**Part c.)**for the integral test I am going to let f(x) = (e

^{(1/x)}-1) and given that the function is positive and e

^{(1/x)}is a decreasing function I may use the integral test.

I will take the limit

_{a→∞}of ∫

_{1}

^{a}f(x)dx and see if it = 0

Immediately my notes suggest I substitute u = 1/x to get ∫-(e

^{u}-1)du/u

^{2}

Here is where i can't quite keep up with the logic my instructor hints about. Apparently I can say that as x→∞, u=0 and it becomes a form of ∫e/x

If it were me I would think that if you claimed that u=0 as x→∞ I would be left with the integral ∫-(eHere is where i can't quite keep up with the logic my instructor hints about. Apparently I can say that as x→∞, u=0 and it becomes a form of ∫e/x

^{P}dx and I can now take some basic facts about the exponential function to determine convergence or divergence without doing the integral.If it were me I would think that if you claimed that u=0 as x→∞ I would be left with the integral ∫-(e

^{0}-1)du/0^2 which is ∫0du/0. Maybe if i had the confidence to say that somehow i am left with ∫-(e^{0}-1)du/u^{2}I could see working with that a little more but i am still at a ∫-(1-1)du/u^{2}which is quite clearly 0 to me.

**To gain some more insight I checked out this link http://www.sosmath.com/calculus/improper/convdiv/convdiv.html on improper integrals and it seems that if i have a 1 to ∞ integral of form 1/x**^{P}it is considered convergent. Maybe if i can just work out a way to make my integral identical to that case i can show that the integral is convergent?

**Any ideas of how I can set up this integral properly to show that it is convergent would be appreciated. I also would like to hear about any corrections that need to be made.**