<s>[INST] Show that $f(n)=\sum_{k=1}^n \frac n {n^2+k^2}$ is strictly increasing 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.
 [/INST] I will supplement @Markus Scheuer's answer by providing a proof of the claim (Corollary 3 of Monotonicity of Riemann Sums by D. Borwein, J. M. Borwein and B. Sims) in a slightly generalized setting.

Lemma. Let $f : [0, 1] \to \mathbb{R}$ and $\lambda \in \mathbb{R}$. Suppose that
  $$F(x) = \lambda f(x) + (1-\lambda) f(1-x) $$
  is concave. If we write $\tau_n = \sum_{k=1}^{n} f\left(\frac{k}{n}\right)\frac{1}{n}$, then we have the following inquality
  $$ \tau_{n+1} \geq \frac{1}{(n+1)^2}f(0) + \frac{n(n+2)}{(n+1)^2} \tau_n. \tag{*} $$

As a corollary, suppose that $f$ and $\lambda$ satisfy the assumption of Lemma. Then


*

*If $f(0)$ is the global maximum of $f$, then $\tau_{n+1} \geq \tau_n$ since $f(0) \geq \tau_n$.

*If $0$ is the unique maximum point of $f$, then we have strict inequality $\tau_{n+1} > \tau_n$.



Proof of Lemma. We may write
$$ \tau_{n+1} = \frac{f(1)}{n+1} + \sum_{k=1}^{n} F\left(\frac{k}{n+1}\right)\frac{1}{n+1}. $$
Now by using the identity $ \frac{k}{n+1} = \frac{k}{n+1}\left(\frac{k-1}{n}\right) + \frac{n+1-k}{n+1} \left( \frac{k}{n} \right) $ together with the concavity of $F$,
$$ F\left(\frac{k}{n+1}\right)
\geq \frac{k}{n+1} F\left(\frac{k-1}{n}\right) + \frac{n+1-k}{n+1} F\left( \frac{k}{n} \right). $$
Plugging this back, we find that
\begin{align*}
\tau_{n+1}
&\geq \frac{f(1)}{n+1} + \sum_{k=1}^{n} \left[ \frac{k}{(n+1)^2} F\left(\frac{k-1}{n}\right) + \frac{n+1-k}{(n+1)^2} F\left( \frac{k}{n} \right) \right] \\
\\
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&= \frac{f(1)}{n+1} + \color{red}{\lambda \sum_{k=1}^{n} \frac{k}{(n+1)^2} f\left(\frac{k-1}{n}\right)} + \color{blue}{(1-\lambda) \sum_{k=1}^{n} \frac{k}{(n+1)^2} f\left(\frac{n-k+1}{n}\right)} \\
&\hspace{4.1em} + \color{blue}{\lambda \sum_{k=1}^{n} \frac{n+1-k}{(n+1)^2} f\left( \frac{k}{n} \right)} + \color{red}{(1-\lambda) \sum_{k=1}^{n} \frac{n+1-k}{(n+1)^2} f\left( \frac{n-k}{n} \right)} \\
\\
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&= \frac{f(1)}{n+1} + \color{red}{\lambda \sum_{k=0}^{n-1} \frac{k+1}{(n+1)^2} f\left(\frac{k}{n}\right)} + \color{blue}{(1-\lambda) \sum_{k=1}^{n} \frac{n+1-k}{(n+1)^2} f\left(\frac{k}{n}\right)} \\
&\hspace{4.1em} + \color{blue}{\lambda \sum_{k=1}^{n} \frac{n+1-k}{(n+1)^2} f\left( \frac{k}{n} \right)} + \color{red}{(1-\lambda) \sum_{k=0}^{n-1} \frac{k+1}{(n+1)^2} f\left( \frac{k}{n} \right)} \\
\\
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&= \frac{f(0)}{(n+1)^2} + \color{red}{\sum_{k=0}^{n-1} \frac{k+1}{(n+1)^2} f\left(\frac{k}{n}\right)} + \color{blue}{\sum_{k=1}^{n} \frac{n+1-k}{(n+1)^2} f\left(\frac{k}{n}\right)} \\
\\
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&= \frac{1}{(n+1)^2}f(0) + \frac{n(n+2)}{(n+1)^2} \tau_n
\end{align*}
as desired. ////
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