$ \sum\limits_{n = 2}^\infty {(2 - e^{\frac{1}{2}} )...(2 - e^{\frac{1}{n}} )}.$
Attempt: let $a_n = (2 - e^{\frac{1}{2}} )...(2 - e^{\frac{1}{n}} )$, therefore we get : $\frac{{a_{n + 1} }}{{a_n }} = 2 - \sqrt[{n + 1}]{e} > 2 - \sqrt[{n + 1}]{{(1 + \frac{1}{n})^{n + 1} }} = 1 - \frac{1}{n}$, which diverges through comparison with the harmonic series.
Is this correct? Can anybody show a better solution?
Your inequality need some proving at least, I knew $ (1+\frac{1}{n})^{\frac{1}{n}} \leq e$ but this one is different, here you have a "-" sign so it's backward and the term is not the same. and if this is true, it's not enough yet. You need an other argument, here comparison is not obvious.
you can find an equivalent of $a_n$ :
$\frac{a_{n}}{a_{n-1}} = 2-e^{\frac{1}{n}} = 1-\frac{1}{n} + O(\frac{1}{n^2}) $
Taking the ln and applying approximation, you get:
$ ln(a_n)-ln(a_{n-1}) = -\frac{1}{n} + O(\frac{1}{n^2}) $
These are divergent series, you can equal the partial sums:
$ ln(a_n) = ln(a_2) -H_n + W_n = K -ln(n) +W_n $ , with $W_n$ the partial sum of a convergent series .
So you get (taking the exp of the relation) : $a_n$ ~ $\frac{B}{n}$ with B a constant. So your series diverges