<s>[INST] Prove the sum $\sum_{n=1}^\infty \frac{\arctan{n}}{n}$ diverges. I must prove, that sum diverges, but...
$$\sum_{n=1}^\infty \frac{\arctan{n}}{n}$$ 
$$\lim_{n \to \infty} \frac{\arctan{n}}{n} = \frac{\pi/2}{\infty} = 0$$ 
$$\lim_{n \to \infty} \frac{ \sqrt[n]{\arctan{n}} }{ \sqrt[n]{n} } = \frac{1}{1} = 1$$
Cauchy's convergence test undefined. 
There is a $E_0\gt0$:
$$\left|\frac{\arctan{n+1}}{n+1} + \frac{\arctan{n+2}}{n+2} + ... + \frac{\arctan{n+p}}{n+p}\right| \ge \frac{\pi}{4}\left|\frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{n+p}\right| \ge \frac{\pi}{4} \frac{p}{n+p} (Let\, p = n) \ge \frac{\pi}{4} \frac{\bcancel{n}}{2\bcancel{n}} =  \frac{\pi}{8} (\sim0.4) \ge E_0 = \frac{1}{8} \gt 0;$$
Now am I correct?
 [/INST] Hint: $\displaystyle\lim_{N\to\infty}\sqrt[^N]{\dfrac\pi2}=1\neq0$. The same holds true if you replace $\dfrac\pi2$ with any other strictly positive 
finite quantity.
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