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?

This answer is essentially the same approach as vesszabo's; I include it just to give another perspective. Set $g(x)=1/(1+x^2)$. Consider the trapezoidal approximation 
$$\int_0^1 g(x) dx \approx \frac{1}{n} \left( \frac{g(0)+g(1/n)}{2} + \frac{g(1/n)+g(2/n)}{n} + \cdots + \frac{g((n-1)/n)+ g(1)}{2} \right)$$
$$=\frac{1}{2n} + f(n) - \frac{1}{4n}.$$
It is well known that the error in the trapezoidal approximation is $-(1-0)^3 M/(12 n^2)$ where $M = g''(x)$ for some $x \in [0,1]$. NEXT PART IS UPDATED, thanks to zyx for pointing out the error. The most positive value of $g''$ is $1/2$, at $x=1$, and the most negative is $-2$ at $x=0$. 
So
$$ \int_0^1 g(x) dx - \left( \frac{1}{2n} + f(n) - \frac{1}{4n} \right) \geq - \frac{1}{24 n^2}$$
or, after a little algebra,
$$f(n) \leq  \frac{\pi}{4} - \frac{1}{4n} + \frac{1}{24 n^2}.$$
Similarly,
$$f(n+1) \geq  \frac{\pi}{4} - \frac{1}{4(n+1)} - \frac{1}{6(n+1)^2}$$
So
$$f(n+1) - f(n) \geq \frac{n^2 + 4n -1}{24 n^2 (n+1)^2}$$
which is $>0$ for $n \geq 1$.
The basic point here is that, when comparing a sum to an integral, your first approach should be Riemann sums (tried by Did, but not good enough), your second should be the trapezoidal rule (used here), and the third should be the full power of Euler-Maclaurin summation (vesszabo's answer).
