Nature of the serie $\sum\prod_{k=2}^n (2-e^{\frac{1}{k}})$ I'd like to determine the nature of the following serie :
$$\sum_{n\ge 2}\prod_{k=2}^n (2-e^{\frac{1}{k}})$$
Let $u_n = \prod_{k=2}^n (2-e^{\frac{1}{k}})$.
So I "have": 
$$\begin{aligned}
\ln(u_n) &= \sum \ln(2-e^{1/k}) 
\\& \approx \sum \ln(1-1/k + o(1/k))\\
& \approx \sum 1/k- o(1/k))\\
& \approx -\ln(n) = \ln(1/n)\end{aligned}$$
So I guess that $u_n = \Theta (1/n)$ and so $\sum u_n$ diverge. But all those calculations are not correct since $k$ is not always "big" and we can not sum "$o$" arbitrarily.
 A: Basic idea: $\ln(2-e^h) = -h +O(h^2)$ as $h\to 0.$ So $\ln(2-e^{1/k}) = -1/k +O(1/k^2).$ Summing the latter from $k=2,\dots, n$ gives $-\ln n + a_n,$ where $a_n$ is bounded. Thus the $n$th term in the sum is $\ge c/n$ for some $c>0,c$ independent of $n.$
A: By partial summation we have $$\sum_{k=2}^{n}\log\left(2-e^{1/k}\right)=n\log\left(2-e^{1/n}\right)-2\log\left(2-e^{1/2}\right)-\int_{2}^{n}\left\lfloor t\right\rfloor \frac{e^{1/t}}{t^{2}\left(2-e^{1/t}\right)}dt
 $$ 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\in\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 $$\lim_{n\rightarrow\infty}\left(2-e^{1/n}\right)^{n}=e^{-1}.
 $$
