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

1. Jul 22, 2012

### learningcalc

1. The problem statement, all variables and given/known data

Determine whether the series converges or diverges

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

2. Relevant equations

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

3. 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.

2. Jul 22, 2012

### Simon Bridge

Welcome to PF;
That is a fine analysis.

3. Jul 22, 2012

### learningcalc

Hi Simon,

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

Thanks,

4. Jul 23, 2012

### 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 :)

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?

5. Jul 23, 2012

### 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

6. Jul 23, 2012

### 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,

7. Jul 23, 2012

### clamtrox

Perhaps you could try comparing it to 1/n.

8. Jul 23, 2012

### learningcalc

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

9. Jul 23, 2012

### Simon Bridge

Oh good that worked: knowing how you were thinking about the problem lets 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 :)