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

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The forum discussion centers on the convergence analysis of the series \(\sum (-1)^n * [e - (1 + 1/n)^n]\). The user initially attempted to apply the Leibniz test for normal convergence, successfully demonstrating that \(\lim a(n) = 0\) as \(n\) approaches infinity. However, they struggled with absolute convergence, concluding that the series diverges based on comparisons with known divergent series. A suggestion was made to simplify the absolute convergence proof by showing that \(|a_n| = e\left(\frac{1}{2n} + O\left(\frac{1}{n^2}\right)\right)\).

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

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

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
[tex]\sum[/tex]|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 1:40am and I just couldn't get LATEX to work, any guides on that matter would be appreciated as well.

Thanks in advance.
Just noticed, this is a self study and I guess I posted it in the wrong place as I assumed. My apologies.
 
Last edited:
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emnethesemn said:
Hi and sorry if I misplaced the thread.
I'm having quite some trouble with analyzing the convergence of the following series :

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

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
[tex]\sum[/tex]|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 1:40am and I just couldn't get LATEX to work, any guides on that matter would be appreciated as well.

Thanks in advance.
Just noticed, this is a self study and I guess I posted it in the wrong place as I assumed. My apologies.

For absolute convergence work with:
[tex]\sum[/tex]|e-(1+1/n)^n|
 
I don't think what you did with the alternating series is correct, and in any case you are over complicating the issue. [itex]\sum (-1)^n a_n[/itex] converges if and only if a) [itex]a_n \to 0[/itex] and b) a_n is eventually decreasing. You already said those 2 facts for this sum in that first section you called "1)", you were done.

For absolute convergence, try showing [tex]| a_n | = e\left( \frac{1}{2n} + O\left(\frac{1}{n^2}\right) \right)[/tex].
 
Last edited:

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