Infinite sequences and series - conv or div - sigma(e^(1/n)/n)

[learningcalc]

Homework Statement

Determine whether the series converges or diverges

Sum from n=1 to infinity ((e^(1/n))/n)

Homework Equations

I am trying to use the limit comparison test to prove it.

The Attempt at a Solution

an = (e^(1/n))/n
bn = e/n

an/bn = e^(1/n)/e

lim n-> infinity an/bn = 1/e

Sum from n=1 to infinity e/n is divergent. (e/n = e(1/n). So sum from n=1 to infinity e/n = e*sum from n=1 to infinity 1/n. sum from n=1 to infinity 1/n is divergent because this is a p-series with n^p where p = 1. For p <= 1 the series is divergent.)

Since either both an and bn are convergent or both are divergent, an must be divergent as bn is divergent.

Thanks for any help.

 

[Simon Bridge]

Welcome to PF;
That is a fine analysis.
What was your question?

 

[learningcalc]

Welcome to PF;
That is a fine analysis.
What was your question?

Hi Simon,

My question is whether or not I am approaching the question correctly and if my answer is justify.

Thanks,

 

[Simon Bridge]

And if I say yes or no, how would you decide that I was right or wrong? I'm just some random entity you met on the internet! At some point you'll be faced with having to solve problems that nobody knows the answers to - how will you yell if you are right or wrong then, with nobody to ask? What you need is some way to do without me :)

Is there anything in particular about your approach you are uncertain about?

Perhaps there is a way to check if the series converges or diverges - say by usig another method or, since you have a computer handy, actually doing the sum to large values of n and seeing if they tend to one value?

 

[learningcalc]

Hi Simon,

Please understand I am not seeking approval nor am I trying to have someone else to just hand me the answer.

I tried to use the comparison test to compare two series: e^(1/n))/n vs. e/n

I know that e^(1/n))/n < e/n. So if e/n is convergent, then e^(1/n))/n must also be convergent. But e/n is divergent, so the comparison test doesn't tell me anything new. I also thought of using the integral test to see what happens, but the integral gets completely out of hand.

The only way I know to test the series is the limit comparison test, and I don't know another method to test it. Yes I did use a computer to test it out, and it is divergent. But I am much more interested to know the fundamentals than just the answer. So I stumble across this forum and thought I might be able to seek a second opinion. I kno

 

[learningcalc]

Personally I don't like to just give the answer when someone asks, but this is not one of those situation. So I know full well your intention Simon.

Regards,

 

[clamtrox]

Perhaps you could try comparing it to 1/n.

 

[learningcalc]

that's brilliant! it's so obvious duh to myself haha! e^0 = 1. Thanks mate.

 

[Simon Bridge]

Oh good that worked: knowing how you were thinking about the problem let's clamtrox, confidently, make a useful suggestion (of course he may have guessed that anyway.) :)

Funnily enough, when I saw the problem statement in your first post I did it the opposite way around - starting by considering the form of the function for large n, concluding it was divergent, then using a more formal method to prove that.

Cheers :)