A tribute (Might Draw Blood)

$\large\text{Given}$ $\large \displaystyle \sum_{k=1}^{\infty}\left({\frac{\displaystyle \sum_{n=1}^{k}nk}{\displaystyle\prod_{n=1}^{k}nk}}\right) = \ x$ $\large\text{First we find}$ $\large {\frac{\displaystyle \sum_{n=1}^{k}nk}{\displaystyle\prod_{n=1}^{k}nk}}$ $\large {\displaystyle \sum_{n=1}^{k}nk} = k+2k+3k......kk$ $\Rightarrow k(1+2+3.....k)$ $\Rightarrow (k)\dfrac{k(k+1)}{2}$ $\Rightarrow (k)\dfrac{k^2+k}{2}$ $\Rightarrow \dfrac{k^3+k^2}{2}$ $\large\text{Now we find}$ $\large{\displaystyle\prod_{n=1}^{k}nk}=k\times2k\times3k......kk$ $\Rightarrow(1\times2\times3......k)\times(k\times k.....k times)$ $\Rightarrow (k!)\times(k^k)$ $\large\text{We can write it as}$ $\large {\frac{\displaystyle \sum_{n=1}^{k}nk}{\displaystyle\prod_{n=1}^{k}nk}}=\dfrac{(k^3+k^2)(1\div2)}{(k!)\times(k^k)}$ $\Rightarrow \dfrac{(k^3+k^2)}{2(k!\times k^k)}$ $\Rightarrow \dfrac{k(k^2+k)}{2k[(k-1)!\times k^{k-1}]}$ $\Rightarrow \dfrac{k^2+k}{2[(k-1)!\times k^{k-1}]}$ $\Rightarrow \dfrac{\dfrac{k}{2}(k+1)}{(k-1)!\times k^{k-1}}$ $\large\text{Now we will equate the equation}$ $\large\displaystyle \sum_{k=1}^{\infty}({\frac{\displaystyle \sum_{n=1}^{k}nk}{\displaystyle\prod_{n=1}^{k}nk}}) = \displaystyle \sum_{k=1}^{\infty}\dfrac{\dfrac{k}{2}(k+1)}{(k-1)!\times k^{k-1}}$ $\large\text{But this is not a geometric progession as the first term is less than the second term and the third term is less than the second term. }$ $\large\text{However,This can be rewritten as}$ $\displaystyle \sum_{k=3}^{\infty}\dfrac{\dfrac{k}{2}(k+1)}{(k-1)!\times k^{k-1}}+a+b$ $\large\text{where a is the first term and b is the second term and the expression will be a geometric progession.}$ $\large\displaystyle \sum_{k=3}^{\infty}\dfrac{\dfrac{k}{2}(k+1)}{(k-1)!\times k^{k-1}}=\dfrac{a_1}{1-\dfrac{a_2}{a_1}}$ $\Rightarrow \dfrac{\dfrac 13}{1 - \dfrac{\dfrac5{192}}{\dfrac13}}$ $\Rightarrow \dfrac{\dfrac{1}{3}}{1-(\dfrac{5}{192}\times3)}$ $\Rightarrow \dfrac{1}{3}\times\dfrac{64}{59}$ $\Rightarrow \dfrac{64}{177}$ $\Rightarrow 0.36158$ $\large a=1$ $\large b=\dfrac32$ $\large\text{Adding all the expressions}$ $\displaystyle \sum_{k=3}^{\infty}\dfrac{\dfrac{k}{2}(k+1)}{(k-1)!\times k^{k-1}}+a+b=0.36158+1+1.5$ $\Rightarrow 2.86158$ $\text{So}$ $\large x=2.86158$ $\text{Hence}$ $\large \displaystyle \sum_{k=1}^{\infty}\left({\frac{\displaystyle \sum_{n=1}^{k}nk}{\displaystyle\prod_{n=1}^{k}nk}}\right) = \ 2.86158$ $\text{This is a awesome problem and solution so please don't feel ashamed to upvote and like it}$

Expand and simplify to obtain

$\sum_{k=1}^{\infty}\left({\frac{\displaystyle \sum_{n=1}^{k}nk}{\displaystyle\prod_{n=1}^{k}nk}}\right)=\sum_{m=1}^{\infty }\dfrac{m+1}{2m^{m-2}m! }.$

It is easy to verify that this sum converges using the ratio test. Then evaluating the sum (using a calculator, or by summing the first few terms since it converges quickly) yields an answer of $\boxed{1.86783}$ rounded to $5$ decimal places.