Show that $sum_n (1-e^{-1/n})$ diverges

Rewrite the integrand as $$ e^{log big(frac{exp(frac{1}{n}) -1}{exp(frac{1}{n})}big)} = e^{log (e^{frac{1}{n}}-1) - frac{1}{n}} > e^{0-frac{1}{n}} $$ since $forall n>0 e^{frac{1}{n}}>1$ hence $log$ is positive. This can be compared to the corresponding integral: $$ int_{1}^{infty}e^{-frac{1}{x}}dx $$ which diverges.

If you want an inequality about the exponential function, $$ e^xge 1+x$$ is your best friend. So $$ e^{frac1n}ge 1+frac1n=frac{n+1}{n}$$ and $$e^{-frac1n}=frac1{e^{frac1n}}le frac n{n+1}=1-frac1{n+1},$$ making $$ 1-e^{-frac1n}ge frac1{n+1}.$$