Putnam 2002, Problem B-3

Show that, for all integers $n > 1$, $\frac{1}{2ne} < \frac{1}{e} - \left(1 - \frac{1}{n}\right)^n < \frac{1}{ne}$.

Solution:
The desired inequalities can be rewritten as  
\[ 1 - \frac{1}{n} < \exp\left( 1 + n \log\left( 1 - \frac{1}{n} \right) \right) < 1 - \frac{1}{2n}. \]  

By taking logarithms, we can rewrite the desired inequalities as  
\[ -\log\left( 1 - \frac{1}{2n} \right) < -1 - n \log\left( 1 - \frac{1}{n} \right) \]  
\[ < -\log\left( 1 - \frac{1}{n} \right). \]  

Rewriting these in terms of the Taylor expansion of \( -\log(1 - x) \), we see that the desired result is also equivalent to  
\[ \sum_{i=1}^\infty \frac{1}{i 2^i n^i} < \sum_{i=1}^\infty \frac{1}{(i + 1) n^i} < \sum_{i=1}^\infty \frac{1}{i n^i}, \]  
which is evident because the inequalities hold term by term.  

Note: David Savitt points out that the upper bound can be improved from \( 1/(ne) \) to \( 2/(3ne) \) with a slightly more complicated argument. (In fact, for any \( c > 1/2 \), one has an upper bound of \( c/(ne) \), but only for \( n \) above a certain bound depending on \( c \).)