I want to show that the function $f:\mathbb N \to \mathbb Q$ defined by$$f(n)=\sum_{k=1}^n \frac n {n^2+k^2}$$
is strictly increasing. Could I have a hint on how to go about this? I have tried playing around with $f(n+1)-f(n)$ but it's all so messy with the different denominators that I'm not sure what to do.
Best Answer
$$ \begin{align} &\sum_{k=1}^{n+1}\frac{n+1}{k^2+(n+1)^2}-\sum_{k=1}^n\frac{n}{k^2+n^2}\tag1\\ &=\frac1{2n+2}+\sum_{k=1}^n\left(\frac{n+1}{k^2+(n+1)^2}-\frac{n}{k^2+n^2}\right)\tag2\\ &=\frac1{2n+2}-\sum_{k=1}^n\frac{n(n+1)-k^2}{\left(k^2+(n+1)^2\right)\left(k^2+n^2\right)}\tag3\\ &=\frac1{2n+2}-\frac1{n(n+1)}\sum_{k=1}^n\frac{1-\frac{k^2}{n(n+1)}}{\left(1+\frac{k^2}{(n+1)^2}\right)\left(1+\frac{k^2}{n^2}\right)}\tag4\\ &\gt\frac1{2n+2}-\frac1{n(n+1)}\sum_{k=1}^n\frac{1-\frac{k(k-1)}{n^2}}{\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{(k-1)^2}{n^2}\right)}\tag5\\ &=\frac1{2n+2}-\frac1{n+1}\sum_{k=1}^n\left[\frac{\frac{k}{n}}{1+\frac{k^2}{n^2}}-\frac{\frac{k-1}{n}}{1+\frac{(k-1)^2}{n^2}}\right]\tag6\\[9pt] &=0\tag7 \end{align} $$ Explanation:
$(2)$: separate the $k=n+1$ term from the left sum and combine the rest
$(3)$: algebra
$(4)$: algebra
$(5)$: $\frac{k}{n+1}\gt\frac{k-1}{n}$ and $\frac{1-xy}{\left(1+x^2\right)\left(1+y^2\right)}$ is strictly decreasing in $x$ when $x,y\in[0,1]$
$(6)$: algebra
$(7)$: telescoping series sums to $\frac12$