Convergence of Series (Harder)

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Homework Help Overview

The discussion revolves around proving the limit of the sequence \((1-\frac{1}{n})^n\) as \(n\) approaches infinity, specifically that it converges to \(\frac{1}{e}\). The context is within the subject area of mathematical analysis, particularly focusing on series and limits.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various approaches, including multiplying by \(e\) and examining the product of \((1-\frac{1}{n})^n\) and \((1+\frac{1}{n})^n\). There are questions about the validity of substituting sequences and the specific convergence to \(\frac{1}{e}\).

Discussion Status

The discussion is active, with participants offering different methods and questioning the assumptions behind the approaches. Some guidance has been provided, but there is no explicit consensus on the best method to prove the convergence.

Contextual Notes

Participants note the challenge of proving convergence specifically to \(\frac{1}{e}\) rather than just convergence in general. There is also mention of potential constraints related to the use of sequences in the proof.

sid9221
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Prove that:
[tex](1-\frac{1}{n})^n \rightarrow \frac{1}{e}[/tex] as [tex]n \to \infty[/tex]

you may use the fact that

[tex](1+\frac{1}{n})^n \rightarrow e[/tex]

I have no idea where to even begin, can someone point me in the right direction ?
 
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Try this: Multiply both sides by e.
[tex](1-\frac{1}{n})^n .e \rightarrow \frac{1}{e} .e[/tex]
[tex](1-\frac{1}{n})^n .(1+\frac{1}{n})^n \rightarrow 1[/tex]
If you can simplify the L.H.S. to get 1 on the R.H.S., then you have proved the convergence of the series.
 
sharks said:
Try this: Multiply both sides by e.
[tex](1-\frac{1}{n})^n .e \rightarrow \frac{1}{e} .e[/tex]
[tex](1-\frac{1}{n})^n .(1+\frac{1}{n})^n \rightarrow 1[/tex]
If you can simplify the L.H.S. to get 1 on the R.H.S., then you have proved the convergence of the series.

Really don't know if you can sub in a sequence like that besides I need to prove it converges to [tex]\frac{1}{e}[/tex] not that it simply converges.
 
[tex](1-\frac{1}{n})^n .(1+\frac{1}{n})^n \rightarrow 1[/tex]
[tex]( (1-\frac{1}{n}).(1+\frac{1}{n})) ^n \rightarrow 1[/tex]
[tex](1-\frac{1}{n}+\frac{1}{n}-\frac{1}{n^2})^n \rightarrow 1[/tex]
[tex](1-\frac{1}{n^2})^n \rightarrow 1[/tex]
[tex]\lim_{n\to \infty}(1-\frac{1}{n^2})^n=1[/tex]
 
sid9221 said:
Prove that:
[tex](1-\frac{1}{n})^n \rightarrow \frac{1}{e}[/tex] as [tex]n \to \infty[/tex]

you may use the fact that

[tex](1+\frac{1}{n})^n \rightarrow e[/tex]

I have no idea where to even begin, can someone point me in the right direction ?
Let ##a_n = \left(1 - \frac{1}{n}\right)^n##. Try determining what ##\log a_n## converges to.
 

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