Prove the sum $\sum_{n=1}^\infty \frac{\arctan{n}}{n}$ diverges.

Question

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?

Answer 1

We have

$$\frac{\arctan n }n\sim_\infty \frac{\pi}{2n}$$ and the harmonic series $\sum\limits\frac1n$ is divergent so the given series is also divergent by the asymptotic comparison.

Answer 2

For large enough $n$, $\text{arctan}(n)\geqslant \pi/4$ and $\sum \frac{\pi/4}{n} $ diverges.

Answer 3

At the end, you have to take the limit as $n$ and $p$ go to infinity, or, if you prefer, the inequality $|S_{n+p}-S_n|\lt \varepsilon$ has to take place for any $n\geqslant N(\varepsilon)$ and $p\geqslant 0$.

Instead, use the inequality $\arctan n\geqslant \pi /4$ for $n\geqslant 1$.

Answer 4

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.

Answer 5

Because $\arctan(x)>1$ when $x>2$, we have that $$\sum_{k=2}^\infty\frac{\arctan(k)}k>\sum_{k=2}^\infty\frac1k$$And the latter sum is the famous harmonic series (minus $1$), which diverges. Therefore, by series comparison, the sum in question diverges.