# A very tough convergence (-1)^n * [e-(1+1/n)^n]

emnethesemn
Hi and sorry if I misplaced the thread.
I'm having quite some trouble with analyzing the convergence of the following series :

## Homework Statement

:[/B]
Determine whether the series is convergent or divergent, absolutely & normally.

$$\sum$$ (-1)^n * [e-(1+1/n)^n]

## The Attempt at a Solution

I had troubles both with absolute and normal convergence.

With normal convergence
I tried Leibniz
1)
lim a(n) = 0 Which is ok { lim e-(1+1/n)^n = 0 } as
(1+1/n)^n rises to e when we let n go to infinity.
2)
a(n)=<v>=a(n+1)
I get to a part in which I have [1+1/n]/[1+1/(n+1)]<=1 I put limes on both sides
and get them to equalize. So I guess normal convergence is fine. But I'm not sure of this.

When testing absolute convergence I figured out that I could state that it's smaller then
$$\sum$$|e|+|(1+1/n)^n|
but these series diverge so I'm nowhere.
I tried integral test but an integral of [(1+1/n)^n]dn with range of 0 to +infinity doesn't seem solvable.

I need to prove it Diverges absolutely, any ideas?
Sorry for all the bad grammar and anything that I left unclear, it's 2:00 am now /yawn and I just couldn't get LATEX to work, any guides on that matter would be appreciated as well.

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estro

Try to think about the root test, it will give you the answer to both of your questions.

gomunkul51

The series is Leibniz convergent as you showed: (limits to zero, (-1)^n and monotonic decreasing).

I'm not sure about absolute convergence..

*** DO you know about the Limit Comparison Test? ;))

note: |e - (1+1/n)^n| = e - (1+1/n)^n
*because e is always greater then (1+1/n)^n and both of them are always positive.

P.S. I don't see how Cauchy's Root Test will help here...

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emnethesemn

Well the only thing I can think of ( that I haven't tried ) is Gauss. I'm not sure if his test would work, and I never really understood that test properly anyway. Cauchy's Root Test doesn't help I think.

Homework Helper

You want to approximate e-(1+1/n)^n. Look at log((1+1/n)^n). Use the taylor series expansion of log(1+x) to get the leading terms.

gomunkul51

Can you explain a bit more Dick? :)
With the approx. I'm not getting anything new, only that the limit is zero...

* On the other hand, I do have a proof using the Limit comparison test and a tricky limit ;)

*** Love those series questions! but that's it for today, I"ll check the thread in 8 hours.

emnethesemn

Gomunkul if you're still here I'd like to see that limit comparison test !!
I don't know how that works.

gomunkul51

http://mathworld.wolfram.com/LimitComparisonTest.html

you need to find a correct series, to compare your series in question with, based on your guess if it is converges or diverges ;)
think about it, check several different series and assumption until you get it.

I'm out...

Homework Helper

Can you explain a bit more Dick? :)
With the approx. I'm not getting anything new, only that the limit is zero...

* On the other hand, I do have a proof using the Limit comparison test and a tricky limit ;)

*** Love those series questions! but that's it for today, I"ll check the thread in 8 hours.

Show n*log(1+1/n)=1-1/(2n)+ terms of higher order in n. Now exponentiate it and subtract it from e. You should be able to show e-(1+1/n)^n=e/(2n)+terms of higher order in n. Conclusion? I suspect you are doing a limit comparison test with 1/n. That's a bit more rigorous. I like it better. I'm just doing the physicists approach.

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gomunkul51

You are right Dick, great Idea.

I need to use Taylor Expansions more!

Basically you got a Harmonic Series in your Taylor expansion of the series in question and we know that those are really fun but do not converge.

** I did use the Harmonic Series for comparison and appropriately the limit I got was e/2 :)