Let $$f(n)=\tan^{-1}\left(\frac{1}{n}\right)+\tan^{-1}\left(\frac{2}{n}\right)+\tan^{-1}\left(\frac{3}{n}\right)+ \cdots +\tan^{-1}\left(\frac{n}{n}\right)$$ where $n\in\mathbb{N}$.
Prove that $f(n)$ increases as $n$ increases.
I first tried to create telescopic sum, but it seems that it is impossible to do so.
I tried to find all $n$ such that $f(n+1)-f(n)≥0$, this would give me terms like $\tan^{-1}(r/n) -\tan^{-1}(r/n+1)$ , which I would simplify using the formula for inverse tangents' sum and difference, but this made it worse.
I used a graphing calculator to find that: as $ x→∞,f(n)→∞$ and hence $f$ doesn't converge.
Then i tried to approximate $f(n);f(n+1)$ as linear functions such that $f(n)<k(n)$ and $f(n+1)>q(n+1)$, such that $q(n+1)≥k(n)$
but I could not find any.
Notice for any $1 \le k \le n$,
$$\frac{k+1}{n+1} - \frac{k}{n} = \frac{n-k}{n(n+1)} \ge 0 \implies \tan^{-1}\frac{k+1}{n+1} \ge \tan^{-1}\frac{k}{n}$$ We have $$f(n+1) = \sum_{k=1}^{n+1}\tan^{-1}\frac{k}{n+1} = \tan^{-1}\frac{1}{n+1} + \sum_{k=1}^n \tan^{-1}\frac{k+1}{n+1} > \sum_{k=1}^n\tan^{-1}\frac{k}{n} = f(n)$$