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Convergence of infinite series (e^(1/n)-1)

  • Thread starter MrMaterial
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  • #1
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Homework Statement


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

Homework Equations


Integral Test
Limit Comparison Test
L'hopital's rule[/B]


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.


 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618

Homework Statement


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

Homework Equations


Integral Test
Limit Comparison Test
L'hopital's rule[/B]


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.

You've got some right ideas and some wrong ideas. When you change variables ##u=1/x## then integral limits change from 1 to infinity to 1 to 0. The convergence of ##1/x^p## on an interval like that depends on the value of ##p##. Read more carefully. And you won't be able to reduce your integral to exactly that form. But the comparison test is your friend. Here's a big hint. Do you know for example that ##e^u \ge 1+u##?
 
  • #3
33
0
img25.gif
is convergent if and only if p<1 Seems relevant here, And yeah I was a little careless forgetting to change my interval of integration after substituting. With the substitution I actually have:

10(eu-1)du/u2 and given this above P condition things are looking divergent to me now.

Dick you mentioned that eu is ≥ 1 + u and that reminds me of something i learned recently about ex that i thought might be of some use here but just don't understand the scope of it well enough to know it is appropriate to use in a situation like this one.

What I learned about ex is that it is the same as 1 + x + x2/2! + x3/3! ... which is a bit outstanding. How do I use this? Your hint though has given me another look at ex, and now I know that I can use it in a comparison test!

Doing more convergence tests in the middle of an integral like this is well out of my comfort zone, but if this is right then that will be a valuable lesson learned.

so now in my integral I have this function f(u) = (eu-1)/u2 and I want to identify a smaller function that diverges so i may perform the comparison test and prove that ∫f(u)du diverges

when I substitute 1+u for eu, I am defining a smaller function because of the relationship inherent of eu we just defined.

0 ≤ (u+1-1)/u2 = 1/u ≤ (eu-1)/u2

I know that 1/u is divergent because it is the harmonic series, but more importantly I know that my new, smaller integral of ∫10du/u is divergent because p = 1. therefore by the ?comparison test? 01(eu-1)/u2 diverges and thus Σn=1 (e(1/n)-1) is a divergent series.

IF this is correct, Thank you Dick for the help but i also wonder why we even bothered with the integral test, does that sort of substitution only work in the scope of an integral? Wouldn't there be another way to show that the series is divergent?
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
img25.gif
is convergent if and only if p<1 Seems relevant here, And yeah I was a little careless forgetting to change my interval of integration after substituting. With the substitution I actually have:

10(eu-1)du/u2 and given this above P condition things are looking divergent to me now.

Dick you mentioned that eu is ≥ 1 + u and that reminds me of something i learned recently about ex that i thought might be of some use here but just don't understand the scope of it well enough to know it is appropriate to use in a situation like this one.

What I learned about ex is that it is the same as 1 + x + x2/2! + x3/3! ... which is a bit outstanding. How do I use this? Your hint though has given me another look at ex, and now I know that I can use it in a comparison test!

Doing more convergence tests in the middle of an integral like this is well out of my comfort zone, but if this is right then that will be a valuable lesson learned.

so now in my integral I have this function f(u) = (eu-1)/u2 and I want to identify a smaller function that diverges so i may perform the comparison test and prove that ∫f(u)du diverges

when I substitute 1+u for eu, I am defining a smaller function because of the relationship inherent of eu we just defined.

0 ≤ (u+1-1)/u2 = 1/u ≤ (eu-1)/u2

I know that 1/u is divergent because it is the harmonic series, but more importantly I know that my new, smaller integral of ∫10du/u is divergent because p = 1. therefore by the ?comparison test? 01(eu-1)/u2 diverges and thus Σn=1 (e(1/n)-1) is a divergent series.

IF this is correct, Thank you Dick for the help but i also wonder why we even bothered with the integral test, does that sort of substitution only work in the scope of an integral? Wouldn't there be another way to show that the series is divergent?
You've got it. I don't see why you couldn't just apply an argument like that to the original series and skip the integral test. But maybe this is all good practice if you have 'fears' of it. If you know the series ##e^x=1+x+x^2/2!+ ...## then it's pretty easy to see that ##e^x \ge 1+x## at least for ##x \ge 0##.
 

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