1. The problem statement, all variables and given/known data Part a.) For a>0 Determine Limn→∞(a1/n-1) Part b.) Now assume a>1 Establish that Σn=1∞(a1/n-1) converges if and only if Σn=1∞(e1/n-1) converges. Part c.) Determine by means of the integral test whether Σn=1∞(e1/n-1) converges 2. Relevant equations Integral Test Limit Comparison Test L'hopital's rule 3. The attempt at a solution Part a.) Looking at the limit: Limn→∞(a1/n-1) I see that a1/n = a0 as n → ∞ Limn→∞(a1/n-1) = Limn→∞(1-1) = 0 Part b.) 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 Limn→∞(a1/n-1)/(e1/n-1) = Limn→∞(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 (a1/n-1) to make anything workable, maybe something can be done with (e1/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 (a1/x-1)/(e1/n-1) = (x2a(1/x)ln(a))/(x2e(1/x)) = ln(a) Limx→∞a1/x/e1/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. Part c.) I do not like the integral test but it's time to face my fears! 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 limita→∞ of ∫1af(x)dx and see if it = 0 Immediately my notes suggest I substitute u = 1/x to get ∫-(eu-1)du/u2 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/xPdx 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 ∫-(e0-1)du/0^2 which is ∫0du/0. Maybe if i had the confidence to say that somehow i am left with ∫-(e0-1)du/u2 I could see working with that a little more but i am still at a ∫-(1-1)du/u2 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/xP 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.