<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] (Partial answer, more some remarks actually.)
These are Riemann sums $f(n)=\frac1n\sum\limits_{k=1}^ng\left(\frac{k}n\right)$ where $g:x\mapsto\frac1{1+x^2}$ on $[0,1]$. 


*

*Since $g$ is decreasing and the Riemann sums are based on the right endpoints of the intervals $\left[\frac{k-1}n,\frac{k}n\right]$, one knows that $f(n)\lt\ell$ for every $n$, where $\ell=\lim\limits_{n\to\infty}f(n)=\int\limits_0^1g=\frac\pi4$.

*The intervals of the partition into $nk$ intervals are subintervals of those of the partition into $n$ intervals hence the same argument shows that $f(n)\lt f(nk)$ for every $n\geqslant1$ and $k\geqslant2$, for example $f(2^n)\lt f(2^{n+1})$ for every $n\geqslant0$.


A proof of the full monotonicity of $(f(n))$, if this property holds, might use more refined properties of $g$ than its monotonicity, perhaps its convexity/concavity properties on $[0,1]$.
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