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

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In summary: With this, you can then use the Divergence Theorem to show that a_n \to 0 as n \to \infty, and that a_n is eventually decreasing.
  • #1
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.
 
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  • #2
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|
 
  • #3
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].
 
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Related to A very tough convergence: (-1)^n * [e-(1+1/n)^n]

1. What is the meaning of (-1)^n in the expression?

The (-1)^n term in this expression indicates that the value of the entire expression alternates between positive and negative depending on the value of n. When n is an even number, (-1)^n is equal to 1, and when n is an odd number, (-1)^n is equal to -1.

2. What is the significance of the e in the expression?

The e in this expression represents the mathematical constant, also known as Euler's number, approximately equal to 2.71828. It is often used in exponential functions and has many important applications in calculus and other areas of mathematics.

3. How does the expression change as n approaches infinity?

As n approaches infinity, the expression becomes increasingly complex and difficult to calculate. However, it can be shown that the value of the expression converges to a specific number, known as the Euler-Mascheroni constant, which is approximately equal to 0.577.

4. What is the purpose of the (1+1/n)^n term?

The (1+1/n)^n term in this expression is known as a "limiting factor." As n approaches infinity, this term approaches 1, and its effects on the overall expression diminish. This allows us to focus on the other, more complex parts of the expression and understand its behavior as n gets very large.

5. What are the real-world applications of this expression?

This expression is often used in mathematical and scientific research to study the convergence of various series and sequences. It also has applications in finance and economics, for example in calculating compound interest or modeling population growth. Additionally, the Euler-Mascheroni constant, which is the limit of this expression, has important applications in number theory and other areas of mathematics.

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