regularization of sum $n \ln(n)$ I was testing out a few summation using my previous descriped methodes when i found an error in my reasoning. I'm really hoping someone could help me out.
The function which i was evaluating was 
$\sum_{n=1}^{\infty} n\ln(n)$ which turns out to be $-\zeta'(-1)$. This made me hope i could confirm my previous summation methode for divergent sums. 
My divergent summtion methode (see previous questions) gives for every $d\geqslant2$:
$$\sum_{n=1}^{\infty}f(dn)-f(n)=\sum_{n=1}^{\infty}\sum_{p=1}^{d-1} f(n)e^{ip\pi2n/d}$$
$$\sum_{k=0}^{\infty} \sum_{n=1}^{\infty}\sum_{p=1}^{d-1} -(d)^{2k} n\ln(d^kn)e^{2i\pi*pn/d}=\sum_{n=1}^{\infty} n \ln(n) \tag 1$$
Fill in $d=2$ cause that's the most easy, gives:
$$\sum_{k=0}^{\infty} \sum_{n=1}^{\infty} -(2)^{2k}k n\ln(2)(-1)^n-(2)^{2k} n\ln(n)(-1)^n$$
$$\sum_{k=0}^{\infty} \sum_{n=1}^{\infty} -(4)^{k} n\ln(n)(-1)^n=\sum_{n=1}^{\infty}  n\ln(n)/3(-1)^n \tag 2$$
$$\sum_{k=0}^{\infty} \sum_{n=1}^{\infty} -(2)^{2k}k n\ln(2)(-1)^n=\sum_{n=1}^{\infty} -\frac{4}{9}n\ln(2)(-1)^n= \frac{1}{9}\ln(2) \tag 3$$
equation (1)=(2)+(3)
$$(\sum_{n=1}^{\infty} (-1)^n*n \ln(n)/3)+\ln(2)/9\approx 0.165421153 \tag 4$$
Now i work out $\lim_{m\to\infty}  \sum_{n=1}^{m} n \ln(n)$
$$\sum_{n=1}^{m} n \ln(n)=m(m+1)\ln(m+1)/2-(m)(m+2)/4+\ln(m+1)/12+error \tag 5$$
It turns out the error is most likely $-1/6+~0.165421153$ so the value above.
The fact i get an expression with the value found above is cool. Actualy i'm close but       
$\textbf{[Question]}$ why the $-1/6$. Did i failed isolating a constant part in my approximation? And if so were did i fail to get it out.     
Ps: since i guessed the formula, it would be nice if someone could confirm the fomula if correct (or false).
 A: [update 2] : the answer is based on a wrong formula; the OP has corrected his formula (1) and thus all the reasoning down to eq (4) are historical now. Possibly I should delete that answer for simply cleaning up things...

Possibly obsolete now:

I begin with that in your post       

$d\geqslant2$:
  $$\sum_{k=0}^{\infty} \sum_{n=1}^{\infty}\sum_{p=1}^{d-1} (d^2)^{k} \ln(d^kn)e^{2i\pi*pn/d}=\sum_{n=1}^{\infty} n \ln(n)$$
  Fill in $d=2$ cause that's the most easy, gives:     

For $d=2$ the innermost loop resolves into one single pass, making the $e^{...}$ the factor $(-1)^n$. Shifting the $(2^2)^k$ out of the then inner loop (where it is a constant) and also separating $\ln(2^k n)$ into $(k \ln2 + \ln n)$ I get the more explicite version:
$$ \sum_{k=0}^{\infty}4^{k} \sum_{n=1}^{\infty} (-1)^n (\ln(n) + k\ln2) \overset{??}=\sum_{n=1}^{\infty} n \ln(n) \tag 1$$ 
The lhs can possibly -with arguments from the concept of divergent summation of alternating series (here Cesaro-summation would suffice)- be separated into
$$ \sum_{k=0}^{\infty}4^{k} \sum_{n=1}^{\infty} (-1)^n \ln(n) + 
    \ln2\sum_{k=0}^{\infty}4^{k} k \sum_{n=1}^{\infty} (-1)^n  \tag 2
$$
But we have $ \sum_{n=1}^{\infty} (-1)^n \ln(n) = \eta'(0) \approx  0.225791352645$, a constant. And also, $ \sum_{n=1}^{\infty} (-1)^n = - \eta(0)=- \frac 12$.
So we can furtherly reduce:
$$  \eta'(0) \sum_{k=0}^{\infty}4^{k} - \frac 12 \ln2\sum_{k=0}^{\infty}4^{k} k \tag 3$$
and after that, using the rational expression of the geometric series I arrive at  :
$$ {  0.225791352645 \over -3} - {\frac 12 \ln2 \cdot 4\over 9} \approx -0.229296491006 \tag 4$$
[update] On the other hand, $$\sum_{n=1}^{\infty} n \ln(n) = -\zeta'(-1) \approx 0.165421143700 $$ $\qquad \qquad \qquad$ (the numerical value obtained by Pari/GP's -zeta'(-1)) 
so it seems, that the identity in (1) does not hold.

So possibly you (and not me...) made some algebraic error because you arrive at


Fill in $d=2$ cause that's the most easy, gives:
  $(\sum_{n=1}^{\infty} (-1)^n*n \ln(n)/3)+\ln(2)/9\approx 0.165421153$          

where you have the cofactor $n$ in the first part and miss the $- 4\cdot \frac 12$ in the second part. (I hope I've got things correct so far...)
A: Using the Euler-Maclaurin Sum Formula, the asymptotic expansion of $\sum\limits_{k=1}^n\log(k)k^{-s}$ is
$$
-\zeta'(s)+\frac{n^{1-s}\log(n)}{1-s}-\frac{n^{1-s}}{(1-s)^2}+\frac{\log(n)}{2n^s}-\frac{s\log(n)}{12n^{1+s}}+\frac{n^{-1-s}}{12}+O\!\left(\frac{\log(n)}{n^{3+s}}\right)\tag{1}
$$
Setting $s=-1$ gives
$$
\sum_{k=1}^n\log(k)k=-\zeta'(-1)+\frac{n^2\log(n)}{2}-\frac{n^2}4+\frac{n\log(n)}2+\frac{\log(n)}{12}+\frac1{12}+O\!\left(\frac1{n^2}\right)\tag{2}
$$
According to $(1)$, the error term should be $O\!\left(\frac{\log(n)}{n^2}\right)$. However, if we compute two more terms at $s=-1$, we get that the error in $(2)$ is $\sim\frac1{720n^2}$.
In any case, the formula in equation $(5)$ from the question is
$$
\frac{n^2\log(n)}2-\frac{n^2}4+\frac{n\log(n)}2+\frac{\log(n)}{12}+\frac14+O\!\left(\frac1{n^3}\right)\tag{3}
$$
However, the correct constant term is
$$
-\zeta'(-1)+\frac1{12}=0.248754477\tag{4}
$$
which is close to $\frac14$, but not exact.
A: I do not know how much this could help you; so, forgive me if I am off-topic. Considering $$S_m=\sum_{n=1}^{m} n\log(n)=\log (H(m))$$ where $$H(m)=\prod_{k=1}^m k^k$$ denotes the hyperfactorial function (defined for positive integers).
For large values of $m$, we can build asymptotics which, limited to second order, give $$S_m=m^2 \left(\frac{1}{2} \log(m)-\frac{1}{4}\right)+\frac{1}{2} m
   \log (m)+\left(\log (A)+\frac{1}{12} \log
   (m)\right)+\frac{1}{720
   m^2}+O\left(\left(\frac{1}{m}\right)^3\right)$$ in which $A$ is  Glaisher's constant $(\approx 1.28243)$.
As you can see, if $m$ is large, you can just forget the last term and have a very good approximation.
To give an idea, $S_{10}\approx 102.082830552$ while the above formula leads to $\approx 102.082830572$. 
Edit
If I use the formula, the maximum error is obtained for $m=1$ and the error is $\approx -1.437 \times 10^{-4}$. For $m=2$, the error becomes $\approx -1.114\times 10^{-5}$; for $m=3$, the error becomes  $\approx -2.327\times 10^{-6}$.
Update
After robjohn's comment, I extended the asymptotics to get 
$$S_m=m^2 \left(\frac{1}{2} \log(m)-\frac{1}{4}\right)+\frac{1}{2} m
   \log (m)+\left(\log (A)+\frac{1}{12} \log
   (m)\right)+\frac{1}{720
   m^2}-\frac{1}{5040
   m^4}+\frac{1}{10080 m^6}-\frac{1}{9504
   m^8}+O\left(\frac{1}{m^{10}}\right)$$
A: @Claude Leibovici I just found your solution, with a little bit different methode, but in the end it's the same and explains the difference in constant part to be found for a general solution. 
When h goes to 0.
$$\sum_{n=1}^m n^{s+h}= \sum_{n=1}^m (n^s ( 1+h\ln(n))$$
$$\sum_{n=1}^m n^{s+h}=\sum_{i=0}^{\infty} a_{s,i} m^{s+1+h-i} $$
At s=1
$$\sum_{n=1}^m n^{1}= m^2/2+m/2+m^0/12+...-1/12$$
The whole point i missed at first was the $m^0$.
$$\sum_{n=1}^m n^{1+h}= m^{2+h}/(2+h)+m^{1+h}/2+m^h*(h+1)/(12)+...+\zeta(-1-h)$$
$$\sum_{n=1}^m n\ln(n)= m^2/2\ln(m)-m^2/4 ＋m/2\ln(m)+1/12ln(n) +(m^0)/12+ ... -\zeta' (-1)$$ 
This $\zeta'(-1)$ is like said before $1/12-ln(A)$. 
so we get  $-1/12+ln(A)$ as the new constant ( and the answer of the regularization of the summation) and the $m^0/12$ is part of the function itself. This is were i got confused in the start. Thanks for all the help, this example helped me a lot.
