Finding $\{a_n\}$ such that its product up to k-th element is the highest. How one can find a distribution $\{a_n\}_{n=1}^{+\infty}$ such that $f_k(a) = \prod\limits_{n=1}^k a_n$ is maximal, i.e. there don't exist another sequence $b$ satisfying the conditions (1), (2) and large enough $K$ such that $\forall k \geqslant K: f_{k}(a) \leqslant f_{k}(b)$.
 A: The thing about
$\dfrac1{n(n+1)}$
is that its sum telescopes.
Suppose
$a_n
=u_n-u_{n+1}
$
where $u_n$ is decreasing
to zero
and
$u_1 = 1$.
Then
$f_n
=\prod_{i=1}^n a_i
=\prod_{i=1}^n (u_i-u_{i=1})
$.
Let's try
$u_n = n^{-c}$
where $c > 0$.
$\begin{array}\\
f_n
&=\prod_{i=1}^n (i^{-c}-(i+1)^{-c})\\
&=(n!)^{-c}\prod_{i=1}^n (1-(1+1/i)^{-c})\\
&\approx (n!)^{-c}\prod_{i=1}^n (1-(1-c/i))\\
&= (n!)^{-c}\prod_{i=1}^n (c/i)\\
&= (n!)^{-c-1}c^n\\
\text{so}\\
\ln f_n
&\approx (-c-1)\ln n!+n \ln c\\
&\approx (-c-1)(n \ln n - n)+n \ln c\\
&= (-c-1)(n \ln n)+n( \ln c+c+1)\\
\end{array}
$
If $c=1$,
which is your example,
this is about
$-2n\ln n+2n$.
If
$c=\frac12$.
this is about
$\frac32(n \ln n)+.8\,n
$.
For small positive $c$,
this is about
$-(1+c)(n \ln n)
$,
which is larger than
your $a_n$.
A more precise result
for $c = \frac12$
is
$\begin{array}\\
\prod_{i=1}^n (\frac1{\sqrt{i}}-\frac1{\sqrt{i+1}})
&=\prod_{i=1}^n \dfrac{\sqrt{i+1}-\sqrt{i}}{\sqrt{i+1}\sqrt{i}}\\
&=\prod_{i=1}^n \dfrac{\sqrt{1+1/i}-1}{\sqrt{i}\sqrt{1+1/i}}\\
&\approx\prod_{i=1}^n \dfrac{1/(2i)}{\sqrt{i}\sqrt{1+1/i}}\\
&=\dfrac1{(n!)^{3/2}2^n}\prod_{i=1}^n \dfrac{1}{\sqrt{1+1/i}}\\
&=\dfrac1{\sqrt{n+1}(n!)^{3/2}2^n}\\
\end{array}
$
and this is larger than your
$\dfrac{1}{(k!)^2 (k+1)}
$.
As $u_n$ goes to zero
more slowly,
the product will get larger.
For example,
suppose
$u_n = \frac{r}{\ln(n+1)}$,
with $r$ appropriately chosen,
say about $\ln 2$.
In this case,
$\begin{array}\\
f_n
&=\prod_{i=1}^n (\frac{r}{\ln(i+1)}-\frac{r}{\ln(i+2)})\\
&=\prod_{i=1}^n (\frac{r(\ln(i+2)-\ln(i+1))}{\ln(i+1)\ln(i+2)})\\
&=\prod_{i=1}^n (\frac{r\ln(1+1/(i+1))}{\ln(i+1)\ln(i+2)})\\
&\approx\prod_{i=1}^n (\frac{r/(i+1)}{\ln(i+1)\ln(i+2)})\\
&=\dfrac{r^n}{(n+1)!}\prod_{i=1}^n (\frac1{\ln(i+1)})^2\\
\end{array}
$
Since
$\ln \prod_{i=1}^n \frac1{\ln(i+1)}
=-\sum_{i=1}^n \ln \ln(i+1)
\approx - n \ln \ln x
$,
$\ln f_n
\approx n \ln r - \ln(n!)- n \ln \ln n
\approx n (\ln r - \ln n + 1- \ln \ln n)
=-n\ln n+ n (\ln r + 1- \ln \ln n)
$
and this is even larger.
This may be analogous to the problem
of finding the
slowest decreasing sequence
whose sum diverges
(with the candidates being
$\frac1{n},
\frac1{n \ln n},
\frac1{n \ln n \ln \ln n},
$
...).
A: There is no maximal sequence. There must be $i,j$ with $a_i\ne a_j$, otherwise $0<a_1=a_i$ for every $i,$ implying  $\sum_ia_i=\infty.$ 
So take $i,j$ with $a_i\ne a_j$. Let $b_n=a_n$ if $i\ne n\ne j$ and let $b_i=b_j=(a_i+a_j)/2.$
Then for all $k\geq \max(i,j)$ we have $f_k(b)/f_k(a)=(a_i+a_j)^2/4a_ia_j>1.$
