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

I’m stuck in showing that $$sum_n (1-e^{-1/n})$$ diverges. The ratio and the root test are inconclusive. Possibly the comparison test is the way to go, but I can’t find a proper bound. I got

$$1-frac{1}{e^{1/n}}>1-frac{1}{2^{1/n}}$$

but still I can’t see here if the last implied series diverges.

Mathematics Asked on November 14, 2021

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.

Answered by Alex on November 14, 2021

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 $$1-e^{-frac1n}ge frac1{n+1}.$$

Answered by Hagen von Eitzen on November 14, 2021