**Problem**: (2002 Putnam - B3) Show that, for all integers $ n > 1 $,

$ \displaystyle \frac{1}{e}-\left(1-\frac{1}{n}\right)^n < \frac{1}{ne} $. [Reworded]

Solution: First of all, this is only one of two inequalities in the problem, but it's the cooler half... yeah. We will rewrite the inequality as

$ \displaystyle 1-\frac{1}{n} < e \left(1-\frac{1}{n}\right)^n $.

Using the fact that $ \displaystyle \lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n = e $ from below, we know

$ \displaystyle e > \left(1+\frac{1}{n}\right)^n $

so

$ \displaystyle e \left(1-\frac{1}{n}\right)^n > \left(1+\frac{1}{n}\right)^n \left(1-\frac{1}{n}\right)^n = \left(1-\frac{1}{n^2}\right)^n > 1-\frac{1}{n} $,

where the last step follows from the binomial expansion or the Bernoulli Inequality. And we're done. QED.

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Comment: Basically, all you really needed was to estimate $ e $ somehow, so the limit form came in handy. The other inequality was showing that the LHS is $ > \frac{1}{2ne} $, but the only solution that I know to that one is taking the $ \log $ and expanding using the Taylor series, which isn't as cool.

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Practice Problem: The left inequality not shown here can also be written as

$ \displaystyle e < \left(1+\frac{1}{n}+\frac{1}{2n(n-1)}\right)^n $.

Can you show this?

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