Thread: easy convergence question

This question seems like it should be so easy, but I just can't quite get there:

Use the fact that $\displaystyle (1 + \frac{1}{n})^n \rightarrow e$ to deduce that $\displaystyle (1 - \frac{1}{n})^n \rightarrow \frac{1}{e}$.

So far this is what I have but I have a feeling i'm going about it the wrong way:
$\displaystyle (\frac{n+1}{n})^n \rightarrow e$
By algebra of limits: $\displaystyle (\frac{n}{n+1})^n \rightarrow \frac{1}{e}$
So $\displaystyle (1- \frac{1}{n+1})^n \rightarrow \frac{1}{e}$.
But from here I can't see how I can get the answer required.

Substitute $\displaystyle n = -m$ and simplify (if you know the limit holds for positive and negative n).

For your solution
$\displaystyle (1- \frac{1}{n+1})^n = (1- \frac{1}{n+1})^{n+1}(1 - \frac{1}{n+1})^{-1} = (1- \frac{1}{n+1})^{n+1}(\frac{n+1}{n})$
Now use the substitution $\displaystyle n = m - 1$

Cheers