Convergence of Series (Harder)

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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.