Does $\lim_{N\rightarrow \infty} \sum_{n = 1}^{N} \frac{1}{(N+1) \ln (N+1) - n \ln n } = 1$? Question:
Find the limit
\begin{equation}
A = \lim_{N\rightarrow \infty} \sum_{n = 1}^{N} \frac{1}{(N+1) \ln (N+1) - n \ln n }
\end{equation}
The series originated from the asymptotic analysis in this question. I can show that it converges. Numerical evaluation in Mathematica suggests that it is close to $1$. 
I am just curious, is it possible to prove one of the following? 


*

*$A > 1$

*$A = 1$

*$A < 1$
Perhaps one can think about the integral
\begin{equation}
\int_0^1 \frac{1}{( 1 + \frac{1}{N} ) \ln (N+1) - x \ln (x N) } dx 
\end{equation}
Update 1:
Let $A_N = \lim_{N\rightarrow \infty} \sum_{n = 1}^{N} \frac{1}{(N+1) \ln (N+1) - n \ln n }$, numerical evaluations up to $10,000$ shows $A_N < 1$

We can also plot the difference $1 - A_N$ as a function of $N$. To check the estimation $1- A_N \sim 0.3 / \ln N $ by one of the comments, we plot $( 1- A_N) \ln N$

update 2
Following Crostul's answer, I did an exercise to prove that the integral is also equal to $1$ in the limit $N \rightarrow \infty$
Relation between $A_N$ and the integral:
\begin{equation}
  A_N = \sum_{n=1}^{N} \frac{1}{N} \frac{1}{ (1 + \frac{1}{N}) \ln ( N+ 1 ) - \frac{n}{N} \ln( \frac{n}{N} N ) }
\end{equation}
so to find $A_{N\rightarrow \infty}$, we may look at the limit
\begin{equation}
\lim_{N \rightarrow  \infty} \int_0^1 \frac{1}{( 1 + \frac{1}{N} ) \ln (N+1) - x \ln (x N) } dx 
\end{equation}
We can simplify the denominator
\begin{equation}
( 1 + \frac{1}{N}  ) \ln ( N+1 ) - x \ln ( xN ) = \ln N ( 1 + \frac{b_N}{N} -x  - \frac{x \ln x}{\ln N} )
\end{equation}
where 
\begin{equation}
b_N = 1 +  ( N + 1 )\frac{\ln (1 + \frac{1}{N})}{\ln N  } = 1 + \frac{1}{\ln N} + o(\frac{1}{N} ) 
\end{equation}
So we study 
\begin{equation}
 I_N = \frac{1}{\ln N} \int_0^1 \frac{1}{ 1 + \frac{b_N}{N} - x - \frac{x \ln x}{\ln N} } dx 
\end{equation}
Now use Crostul's relaxing trick
Define $f(x) = x + \frac{x \ln x}{ \ln N}$, the denominator is $f(1 + \frac{1}{N} ) - f(x)$. On one hand we have
\begin{equation}
f(1 + \frac{1}{N} ) - f(x) \ge  f(1 + \frac{1}{N} ) - x \ge 1 + \frac{1}{N}  - x 
\end{equation}
where the second inequality holds for large enough $N$. 
On the other hand, by mean value theorem 
\begin{equation}
  \frac{f( 1 + \frac{1}{N} ) - f(x) }{1 + \frac{1}{N} - x} = f'( y) \le f'( 1 + \frac{1}{N} ) = 1 + \frac{1 + \ln (1 + \frac{1}{N})}{ \ln N }
\end{equation}
Hence 
\begin{equation}
 \frac{1}{1 + \frac{1 + \ln (1 + \frac{1}{N})}{ \ln N }} \int_0^1 \frac{1}{1 + \frac{1}{N } - x } dx  \le     I_N  \ln N \le \int_0^1 \frac{1}{1 + \frac{1}{N } - x } dx 
\end{equation}
Both integrals on the left and right are $\ln N$. Therefore
\begin{equation}
 \frac{1}{1 + \frac{1 + \ln (1 + \frac{1}{N})}{ \ln N }}  \le I_N \le 1 
\end{equation}
Taking the limit $N\rightarrow$, we have $\lim_{N\rightarrow \infty } I_N = 1$ .

I think if you set $b_{k+1} = a_{k+1} - a_k$, where $a_k = k \log k$ you could rewrite the denominator as $b_{k+1} = \log (1+\frac{1}{k})^k + \log (k+1) < 1+ \log (k+1) < 2 \log (k+1) < 2(k+1)$ hence the sum you have is $S_n = \sum_{k=1}^{n}\frac{1}{b_k} > \frac{1}{2}\sum_{k=1}^{n} \frac{1}{k+1}$ which diverges.
