Discuss convergence of the series $\sum \left\{e-(1+\frac{1}{k})^k \right\}$

Question

This is probably a pretty elementary problem but I was struggling to show the convergence/divergence of the series $$\sum_{k=1}^{\infty} \left\{ e-(1+\frac{1}{k})^k \right\}.$$

Any help appreciated!

Answer 1

Estimate the error term. It turns out to be bounded below by order $1/k$, so the sum diverges by the comparison test.

Indeed, for $x\to 0^+$, we have $$ \frac{(1+x)^{1/x}-e}{x}\to-\frac{1}{2}e\tag{$\dagger$} $$ so $e-(1+\frac1k)^k\approx\frac{e}{2k}$.

Addendum One way to show the limit ($\dagger$) is to look at $$ \log(1+x)^{1/x}=x^{-1}(x-\frac12 x^2+\dots)=1-\frac12 x+\dots $$ so $(1+x)^{1/x}=\exp(1-\frac12x+\dots)=e\cdot\exp(-\frac12 x+\dots)=e-\frac12 ex+O(x^2)$.