Does $\lim\limits_{n \to +∞} \sum_{k=1}^n \frac{n\cdot \ln (k)}{n^2+k^2}$ diverge? Does the limit of this summation diverge?
$$\lim\limits_{n \to +∞} \sum_{k=1}^n \frac{n\cdot \ln (k)}{n^2+k^2}$$
Thanks!
 A: $$\sum\limits_{k = 1}^n {\frac{{n\ln k}}{{{n^2} + {k^2}}}}  = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{\ln k}}{{1 + {{(\frac{k}{n})}^2}}}}  = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{\ln (\frac{k}{n})}}{{1 + {{(\frac{k}{n})}^2}}}}  + \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{\ln n}}{{1 + {{(\frac{k}{n})}^2}}}} $$
When $n \to \infty $, the first term is the definite integral $$\int_0^1 {\frac{{\ln x}}{{1 + {x^2}}}dx} $$
which obviously converge.
To investigate the second term, we note that
$$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + {{(\frac{k}{n})}^2}}}} } \right] = \int_0^1 {\frac{1}{{1 + {x^2}}}dx} $$
Hence your sequence diverges.

PS:
This method can be used to obtain estimate on your sequence, using
$$\frac{\pi }{4} - \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + {{(\frac{k}{n})}^2}}}}  = \frac{1}{{4n}} + o(\frac{1}{n})$$
we have
$$\sum\limits_{k = 1}^n {\frac{{n\ln k}}{{{n^2} + {k^2}}}}  =  - G + \frac{\pi }{4}\ln n - \frac{{\ln n}}{{4n}} + o(\frac{{\ln n}}{n})$$
with $G=-\int_0^1 {\frac{{\ln x}}{{1 + {x^2}}}dx}$ is the Catalan constant.
A: $$\sum_{k\leq n}\frac{n\log\left(k\right)}{n^{2}+k^{2}}\geq\frac{1}{2n}\sum_{k\leq n}\log\left(k\right)
 $$ and by partial summation we can see that $$\sum_{k\leq n}\log\left(k\right)=n\log\left(n\right)-\int_{1}^{n}\frac{\left\lfloor t\right\rfloor }{t}dt\geq n\log\left(n\right)-n+1
 $$ where $\left\lfloor t\right\rfloor $ is the floor function, so the sequence diverges.
A: The summand is $$\frac{n \ln{k}}{n^2+k^2} \geq \frac{n \ln{k}}{2n^2} = \frac{\ln{k}}{2n}$$
Therefore the sum, if it exists, is greater than or equal to
$$\frac{1}{2n} \sum_{k=1}^n \ln{k} = \frac{1}{2n} \log{n!}$$
This diverges, by L'Hôpital's rule, expressing $n!$ as $\Gamma(n+1)$.
