<s>[INST] How prove this $f(n)\le f(n+1)$ where $f(n)=\sum_{k=1}^{n}\frac{n}{n^2+k^2}$ let $$f(n)=\sum_{k=1}^{n}\dfrac{n}{n^2+k^2}$$
prove or disprove 
$$f(n)\le f(n+1)$$
this inequality is found when I deal this  follow limit: 
$$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{1+(k/n)^2}=\int_{0}^{1}\dfrac{1}{1+x^2}dx=\dfrac{\pi}{4}$$
But I can't prove 
$$f(n)\le f(n+1)$$
since
$$f(n+1)=\dfrac{n+1}{(n+1)^2+1}+\dfrac{n+1}{(n+1)^2+2^2}+\cdots+\dfrac{n+1}{(n+1)^2+n^2}+\dfrac{n+1}{(n+1)^2+(n+1)^2}$$
$$f(n)=\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2^2}+\cdots+\dfrac{n}{n^2+n^2}$$
so
$$f(n+1)-f(n)=\left(\dfrac{n+1}{(n+1)^2+1}-\dfrac{n}{n^2+1}\right)+\left(\dfrac{n+1}{(n+1)^2+2^2}-\dfrac{n}{n^2+2^2}\right)+\cdots+\left(\dfrac{n+1}{(n+1)^2+n^2}-\dfrac{n}{n^2+n^2}\right)+\dfrac{1}{2(n+1)}$$
so
$$f(n+1)-f(n)=\sum_{k=1}^{n}\dfrac{k^2-n^2-n}{(k^2+n^2)((n+1)^2+k^2)}+\dfrac{1}{2(n+1)}$$
This problem is my  found it,can you help to solve this problem?
 [/INST] This is a comment that’s too long for the usual format. Let $g(x)=\dfrac{1}{1+x^2}$. Let us generalize the original question by putting for any $y \geq  x \geq 0$,
$$
\begin{array}{lcl}
u_n(x,y)&=&\frac{1}{n}\sum_{k=1}^{n} g\left(x+\frac{k}{n}(y-x)\right) \\
f_n(x,y)&=& u_{n+1}(x,y)-u_n(x,y)  \\
\end{array}
$$
Conjecture $f_n(x,y)$ can be written 
$$
f_n(x,y)=(y-x)h_n(x,y)\prod_{k=1}^n g\left(x+\frac{k}{n}(y-x)\right)g\left(x+\frac{k}{n+1}(y-x)\right)
$$
where $h_n$ is a polynomial with nonnegative coefficients in $x,y$.  
I have checked this conjecture up to $n=10$, and it clearly is stronger
than the original claim.
</s>