series divergence $\sum_{k=1}^{\infty}\sqrt{\tan^{-1}(\frac{1}{k^2})}$ Using the limit compression test
$$\lim_{k\to \infty} \frac{\sqrt{\tan^{-1}(\frac{1}{k^2})}}{\frac{1}{k}}=\infty$$    
But is there a smaller series that diverge? so we can use it to prove that $\sqrt{\tan^{-1}(\frac{1}{k^2})}$ diverge?
 A: From
$$
\arctan x > \frac{x}2,\qquad x\in (0,1),
$$ you get
$$
\arctan \frac1{k^2} > \frac12\:\frac1{k^2},\qquad k=1,2,\ldots
$$ and
$$
\sqrt{\arctan \frac1{k^2}} > \frac{\sqrt{2}}2\:\frac1{k},\qquad k=1,2,\ldots
$$ thus, for $N \geq1$,

$$
\sum_{k=1}^N\sqrt{\arctan \frac1{k^2}} > \frac{\sqrt{2}}2\:\sum_{k=1}^N\frac1{k}
$$ 

and the initial series is divergent as is the harmonic series.
A: $$\lim_{x\to\infty}\frac{\sqrt{\arctan\frac1{x^2}}}{\frac1{x}}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{-\frac2{x^3}\frac1{1+\frac1{x^4}}\frac1{2\sqrt{\arctan\frac1{x^2}}}}{-\frac1{x^2}}=-\lim_{x\to\infty}\frac{x^4}{(x^4+1)2x\sqrt{\arctan\frac1{x^2}}}=1$$
and thus the series $\;\sum\sqrt{\arctan\frac1{n^2}}\;$ converges iff the series $\;\sum\frac1{n}\;$ converges.
A: hint you can use $arctan(1/k^2)\approx 1/k^2$ for small k which can be proved by Taylor series.
A: If you are looking for a series that diverges and has smaller terms than $a_k=1/k$, try $b_k=1/(k\ln k)$, which in series from $k=2$ diverges due to the integral test. 
