Does $\sum\sqrt{n\arctan(1/n^3)}$ converge?

Question

Does the following series converge? Why?

$$\sum\sqrt{n\arctan(1/n^3)}$$

The only way that comes to me is to look at the series:

$$\arctan(x)=\sum\frac{(-1)^k}{2k+1}x^{2k+1}\Longrightarrow \arctan\left(\frac{1}{n^3}\right)=\sum\frac{(-1)^k}{2k+1}n^{-6k-3}$$

As a result (and if I got the computation right), we have:

$$n\arctan(1/n^3)=\sum\frac{(-1)^k}{2k+1}n^{-6k-2}\approx \frac{1}{n^2}-\frac{1}{3n^8}+\cdots$$

Thus, by taking square root, the leading term "becomes" $\frac{1}{n}$, so the series diverges.

I know the answer is "diverge" but my argument (at least the last part?) is far from being rigorous, if not wrong. Any thought/suggestion/better ways of doing this?

Answer 1

You have the right idea. If you want to make it a bit more rigorous, note that if $0\leq x\leq 1$ then $$ \arctan(x)=\int_0^x\frac{dt}{1+t^2}\geq \frac{1}{2}\int_0^x\;dt=\frac{x}{2}$$ hence $$ n\arctan\Big(\frac{1}{n^3}\Big)\geq \frac{1}{2n^2} $$ and therefore $$ \sum_{n=1}^{\infty}\sqrt{n\arctan\Big(\frac{1}{n^3}\Big)}\geq \frac{1}{\sqrt{2}}\sum_{n=1}^{\infty}\frac{1}{n}=\infty$$

Answer 2

$$\lim_{n\to\infty}\frac{\sqrt{n\arctan\dfrac1{n^3}}}{\dfrac1n}=\lim_{n\to\infty}\sqrt{n^3\arctan\frac1{n^3}}=\sqrt{\lim_{x\to0^+}\frac{\arctan x}x}=1$$

and the series diverges like the harmonic series.

Answer 3

Another approach, less simple than carmichael561's answer.

As you did, starting with Taylor series $$\tan ^{-1}\left(\frac{1}{n^3}\right)=\frac{1}{n^3}-\frac{1}{3 n^9}+\frac{1}{5 n^{15}}+O\left(\frac{1}{n^{21}}\right)$$ $$n\tan ^{-1}\left(\frac{1}{n^3}\right)=\frac{1}{n^2}-\frac{1}{3 n^8}+\frac{1}{5 n^{14}}+O\left(\frac{1}{n^{20}}\right)$$ $$n\tan ^{-1}\left(\frac{1}{n^3}\right)=\frac{1}{n^2}\left(1-\frac{1}{3 n^6}+\frac{1}{5 n^{12}}+O\left(\frac{1}{n^{18}} \right)\right)$$ $$\sqrt{n\tan ^{-1}\left(\frac{1}{n^3}\right)}=\frac 1 n \sqrt{1-\frac{1}{3 n^6}+\frac{1}{5 n^{12}}+O\left(\frac{1}{n^{18}} \right) }$$ Now using the generalized binomial therorem $$\sqrt{1-\frac{1}{3 n^6}+\frac{1}{5 n^{12}}+O\left(\frac{1}{n^{18}} \right) }=1-\frac{1}{6 n^6}+\frac{31}{360 n^{12}}+O\left(\frac{1}{n^{18}}\right)$$ which makes $$\sqrt{n\tan ^{-1}\left(\frac{1}{n^3}\right)}=\frac 1 n-\frac{1}{6 n^7}+\frac{31}{360 n^{13}}+O\left(\frac{1}{n^{19}}\right)$$ Using this last results, computing the partial sums and introducing the generalized harmonic numbers $$S_p=\sum_{n=1}^p \sqrt{n\tan ^{-1}\left(\frac{1}{n^3}\right)}=H_p-\frac{H_p^{(7)}}{6}+\frac{31 H_p^{(13)}}{360}+\cdots$$ Now, using the asymptotics of harmonic numbers $$S_p=\log \left(p\right)+\left(\gamma +\frac{\psi ^{(6)}(1)}{4320}-\frac{31 \psi ^{(12)}(1)}{172440576000}\right)+\frac{1}{2 p}+O\left(\frac{1}{p^2}\right)$$ $$S_p\approx 0.495279+\log \left(p\right)+\frac{1}{2 p}+O\left(\frac{1}{p^2}\right)$$ which, for sure, shows the infinite limit.

For illustration purposes, let me show how good is the approximating formula for partial sums $$\left( \begin{array}{cccc} p & \text{exact} & \text{approximation} & \text{difference} \\ 10 & 2.81381 & 2.84786 & -0.03405 \\ 20 & 3.48259 & 3.51601 & -0.03343 \\ 30 & 3.87983 & 3.91314 & -0.03331 \\ 40 & 4.16339 & 4.19666 & -0.03327 \\ 50 & 4.38405 & 4.41730 & -0.03325 \\ 60 & 4.56472 & 4.59796 & -0.03324 \\ 70 & 4.71768 & 4.75092 & -0.03323 \\ 80 & 4.85033 & 4.88356 & -0.03323 \\ 90 & 4.96742 & 5.00064 & -0.03323 \\ 100 & 5.07222 & 5.10545 & -0.03323 \\ 200 & 5.76288 & 5.79610 & -0.03322 \\ 300 & 6.16751 & 6.20073 & -0.03322 \\ 400 & 6.45478 & 6.48799 & -0.03322 \\ 500 & 6.67767 & 6.71089 & -0.03322 \\ 600 & 6.85982 & 6.89304 & -0.03322 \\ 700 & 7.01386 & 7.04707 & -0.03322 \\ 800 & 7.14730 & 7.18052 & -0.03322 \\ 900 & 7.26501 & 7.29823 & -0.03322 \\ 1000 & 7.37032 & 7.40353 & -0.03322 \end{array} \right)$$