Convergence of $\sum_{n=0}^\infty (-1)^n (e-(1+\frac{1}{n})^n)$ Does $\sum_{n=0}^\infty (-1)^n (e-(1+\frac{1}{n})^n)$ converge absolutely, conditionally, or diverge? Attempt: Yes, by the ratio test we have $$ \lim_{n \to \infty} \left| \frac{(-1)^{n+1} (e-(1+\frac{1}{n+1})^{n+1})}{(-1)^n (e-(1+\frac{1}{n})^n)}\right| = \lim_{n \to \infty} \left| \frac{-e+(1+\frac{1}{n+1})^{n+1})}{e-(1+\frac{1}{n})^n}\right| = \left| \frac{ -e + e }{ e - e} \right| = 0$$ So the series converges absolutely.

Using Taylor series we get:

$$\left(1+\frac1n\right)^n=\exp\left(n\ln\left(1+\frac1n\right)\right)=\exp\left(1-\frac1{2n}+\mathcal O\left(\frac1{n^2}\right)\right)$$ so the given general term of the series denoted by $u_n$ is $$u_n=\frac{(-1)^n}{2n}+\mathcal O\left(\frac1{n^2}\right)$$ so the series $\sum u_n$ is convergent since it's sum of the convergent series $\sum \frac{(-1)^n}{2n}$ using the Leibniz theorem and the convergent series $\sum \mathcal O\left(\frac1{n^2}\right)$ by comparison with the Riemann series.

**Remark** This series isn't absolutely convergent since $$|u_n|\sim_\infty\frac1{2n}$$ and all we know that the harmonic series $\sum\frac1n$ is divergent.