Convergence of Series (Harder)

[sid9221]

Prove that:
(1-\frac{1}{n})^n \rightarrow \frac{1}{e} as n \to \infty

you may use the fact that

(1+\frac{1}{n})^n \rightarrow e

I have no idea where to even begin, can someone point me in the right direction ?

 

[DryRun]

Try this: Multiply both sides by e.
(1-\frac{1}{n})^n .e \rightarrow \frac{1}{e} .e
(1-\frac{1}{n})^n .(1+\frac{1}{n})^n \rightarrow 1
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.

 

[sid9221]

Try this: Multiply both sides by e.
(1-\frac{1}{n})^n .e \rightarrow \frac{1}{e} .e
(1-\frac{1}{n})^n .(1+\frac{1}{n})^n \rightarrow 1
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 \frac{1}{e} not that it simply converges.

 

[DryRun]

(1-\frac{1}{n})^n .(1+\frac{1}{n})^n \rightarrow 1
( (1-\frac{1}{n}).(1+\frac{1}{n})) ^n \rightarrow 1
(1-\frac{1}{n}+\frac{1}{n}-\frac{1}{n^2})^n \rightarrow 1
(1-\frac{1}{n^2})^n \rightarrow 1
\lim_{n\to \infty}(1-\frac{1}{n^2})^n=1

 

[vela]

Prove that:
(1-\frac{1}{n})^n \rightarrow \frac{1}{e} as n \to \infty

you may use the fact that

(1+\frac{1}{n})^n \rightarrow e

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.