# On the proof that $\sum n\arctan(n)$ diverges

$\sum_{n=1}^{\infty} n\arctan(n)$ converge or diverge?

I know this diverges because $$\lim_{n\to\infty} n\arctan(n) = \infty \neq 0$$ So by divergence test this series clearly diverges. My question is, can I just say this and end the question?

• Yes, if the limit of the summand diverges or converges to anything other than zero then the sum certainly diverges (as you noted). So yes, you can just say this and end the question. – Grant B. Mar 29 '17 at 5:08

Yes, that's enough. It should be a rule in your notes that if $a_n$ doesn't converge to $0$, then $\sum a_n$ diverges.

This is too long for a comment. It has been added for your curiosity.

As Omnomnomnom already answered, what you wrote is enough to prove the divergence.

Just in case you would like to go further, considering $$S=\sum_{n=0}^\infty n\tan^{-1}(n)=\sum_{n=0}^p n\tan^{-1}(n)+\sum_{n=p+1}^\infty n\tan^{-1}(n)$$ for any $p$ we can use $$\tan ^{-1}(p)+\tan ^{-1}\left(\frac{1}{p}\right)=\frac \pi 2$$ and, for large $p$ $$\tan ^{-1}\left(\frac{1}{p}\right)=\frac{1}{p}+O\left(\frac{1}{p^3}\right)$$ which makes $$\sum_{n=p+1}^\infty n\tan^{-1}(n)\approx\sum_{n=p+1}^\infty n\left(\frac \pi 2-\frac 1 n\right)=\frac \pi 2\sum_{n=p+1}^\infty n-\sum_{n=p+1}^\infty 1$$

In fact, if you consider the partial sum $$S_p=\sum_{n=0}^p n\tan^{-1}(n)$$ using series, you could show that $$S_p\approx \frac{\pi }{4}p^2+\left(\frac{\pi }{4}-1\right) p+c-\frac 1{3p}+\cdots$$ where $c$ is a small number.

Edit

Just for the fun, let $p=10^k$. The table below reports the "exact" value and what gives the above approximation limited to the two first terms $$\left( \begin{array}{ccc} k & \text{"exact"} & \text{approximation} & \Delta\\ 1 & 76.7773 & 76.3938 & 0.38355 \\ 2 & 7832.93 & 7832.52 & 0.411898 \\ 3 & 785184. & 785184. & 0.414881 \\ 4 & 7.85377\times 10^7 & 7.85377\times 10^7 & 0.415181 \\ 5 & 7.85396\times 10^9 & 7.85396\times 10^9 & 0.415211 \\ 6 & 7.85398\times 10^{11} & 7.85398\times 10^{11} & 0.415405 \end{array} \right)$$

• For what it's worth, I think this is great. – Omnomnomnom Mar 29 '17 at 6:45
• @Omnomnomnom. Be sure I appreciate ! – Claude Leibovici Mar 29 '17 at 7:00