Does a limit of zero guarantee series convergence?

[flyingpig]

Homework Statement

Find if this converge or diverges

$$\sum_{n=1}^{\infty} ne^{-n}$$

Solution

$$\lim_{n\rightarrow \infty} ne^{-n} = 0$$ using L'Hopital's rule. Thus $$\sum_{n=1}^{\infty} ne^{-n}$$ converges



My Problem with this

WHAT?? I THOUGHT JUST BECAUSE THE LIMIT GOES TO 0, IT CONCLUDES NOTHING ABOUT IT'S CONVERGENCE!?? WHY ARE THEY CONTRADICTING THEMSELVES? I feel like suing Mr.Stewart for making my Calculus life difficult

 

[snipez90]

Eh if you think about it the author doesn't really have an incentive to write correct solutions. You are right of course.

 

[gb7nash]

What in the hell? What's the name of this book? If that is the answer and justification for that question, that's embarrassing.

 

[flyingpig]

James Stewart, 6e, and they are coming out with a 7e, I bet you they are going to make the same mistake in the new edition too.

 

[lurflurf]

Stewart is garbage, He most likely had nothing to do with that solution and does not care about its accuracy. This is however easily fixed.
consider sum f(n)
lim f(n)=0
is a necessary condition for convergence, but not a sufficient one
lim f(n)+f(n+1)+...+f(n+m-1)+f(n+m)=0
is necessary and sufficient
here f(n)=n e^-n
0<f(n)+f(n+1)+...+f(n+m-1)+f(n+m)<m(1+m/n)f(n)
lim m(1+m/n)f(n)=lim f(n)=0
lim f(n)=0->convergence
as asserted.
So really the statement is true, but misleading if omited details are not infered.

 

[lurflurf]

James Stewart, 6e, and they are coming out with a 7e, I bet you they are going to make the same mistake in the new edition too.

That is Stewarts version of improvement, retain existing errors and add new ones.

 

[flyingpig]

It's funny because I think in somewhere in chapter 9, it still says "Work = force x distance" instead of "work = force x displacement"

 

[paragrafo]

err, yeah its definitely not clear, but to ease your preocupation use comparison with 1/n^2 and use l'hopital several times