Naturaleza de la serie $\sum\prod_{k=2}^n (2-e^{\frac{1}{k}})$

Question

Me gustaría determinar la naturaleza de la serie siguiente:

$$\sum_{n\ge 2}\prod_{k=2}^n (2-e^{\frac{1}{k}})$$

Que $u_n = \prod_{k=2}^n (2-e^{\frac{1}{k}})$.

Para "Tener": $$\begin{aligned} \ln(u_n) &= \sum \ln(2-e^{1/k}) \\& \approx \sum \ln(1-1/k + o(1/k))\\ & \approx \sum 1/k- o(1/k))\\ & \approx -\ln(n) = \ln(1/n)\end{alineado}$$ así que supongo que el $u_n = \Theta (1/n)$ y $\sum u_n$ divergen. Pero todos esos cálculos no son correctos ya que $k$ no siempre es "grande" y no podemos sumar '$o$' arbitrariamente.

Answer 1

Idea básica: $\ln(2-e^h) = -h +O(h^2)$ $h\to 0.$ para $\ln(2-e^{1/k}) = -1/k +O(1/k^2).$ Resumen Este último de $k=2,\dots, n$ da $-\ln n + a_n,$ donde $a_n$ es limitado. Así el término de th de $n$ en la suma es $\ge c/n$ $c>0,c$ $n.$ independiente

Answer 2

Por parcial de la suma tenemos $$\sum_{k=2}^{n}\log\left(2-e^{1/k}\right)=n\log\left(2-e^{1/n}\right)-2\log\left(2-e^{1/2}\right)-\int_{2}^{n}\left\lfloor t\right\rfloor \frac{e^{1/t}}{t^{2}\left(2-e^{1/t}\right)}dt$$ where $\left\lfloor t\right\rfloor =t-\left\{ t\right\}$ is the floor function and $\left\{ t\right\}$ is the sawthoot function. Then $$=n\log\left(2-e^{1/n}\right)-2\log\left(2-e^{1/2}\right)-\int_{2}^{n}\frac{e^{1/t}}{t\left(2-e^{1/t}\right)}dt+\int_{2}^{n}\left\{ t\right\} \frac{e^{1/t}}{t^{2}\left(2-e^{1/t}\right)}dt$$ and using $\left\{ t\right\} \geq0$ we get $$\geq n\log\left(2-e^{1/n}\right)-\int_{2}^{n}\frac{e^{1/t}}{t\left(2-e^{1/t}\right)}dt.$$ Now put in the integral $e^{1/t}=u$. We get $$\sum_{k=2}^{n}\log\left(2-e^{1/k}\right)\geq n\log\left(2-e^{1/n}\right)-\int_{e^{1/n}}^{e^{1/2}}\frac{1}{\left(2-u\right)\log\left(u\right)}du$$ and now we observe that, if $n$ is sufficiently large, $\frac{1}{\left(2-u\right)\log\left(u\right)}$ get a maximum at $e^{1/2}$ for $u\en\left[e^{1/2},e^{1/n}\right]$, so $$\sum_{k=2}^{n}\log\left(2-e^{1/k}\right)\geq n\log\left(2-e^{1/n}\right)-\frac{2}{\left(2-e^{1/2}\right)}\left(e^{1/2}-e^{1/n}\right)=n\log\left(2-e^{1/n}\right)+O\left(1\right)$$ so finally we get $$\sum_{n\geq2}\prod_{k=2}^{n}\left(2-e^{1/k}\right)\geq\sum_{n\geq2}\left(2-e^{1/n}\right)^{n}e^{O\left(1\right)}$$ and the series diverge because $$\lim_{n\rightarrow\infty}\left(2-e^{1/n}\right)^{n}=e^{-1}.$$