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robjohn
robjohn
04:44
$f$ is defined as
$$
f(x)=\sum_{k=1}^\infty\frac{\arctan(kx)}{k^2}\tag1
$$
For $x\le1$, the Mean Value Theorem says that there is a $h\in(0,x)$ so that
$$
\begin{align}
\frac{f(x)-f(0)}x
&=f'(h)\tag{2a}\\
&=\sum_{k=1}^\infty\frac1{k\left(1+k^2h^2\right)}\tag{2b}\\
&\ge\sum_{k=1}^{\lfloor1/h\rfloor}\frac1{2k}\tag{2c}\\
&\ge\frac12\log\left(\lfloor1/h\rfloor\right)\tag{2d}\\
&\ge\frac12\log\left(\lfloor1/x\rfloor\right)\tag{2e}
\end{align}
$$
This means that
$$
\begin{align}
f'(0)
&=\lim_{x\to0}\frac{f(x)-f(0)}x\tag{3a}\\
 
6 hours later…
sunny
sunny
10:32
Thank you @robjohn! Helps a lot seeing another solution. I see, so you are using the inequality $\log n<H_n$, where $H_n$ is the partial sum of the harmonic series. Neat! Is $x\leq 1$ purposefully chosen for some reason? All I can see is that it really needs to be positive in $(2e)$, otherwise $\log$ isn't defined. Moreover, maybe the limit in $(3a)$ then becomes only a right-hand side limit, or?
Wait, I see why you chose $x\leq 1$, so that $\lfloor 1/x \rfloor >0$. Still, $x$ is positive here, or?
robjohn
robjohn
11:35
@sunny Yes, $0\lt x\lt1$ is chosen here, but as we are looking at $x\to0^+$, that is not a problem. Since $f'$ is even, the result gives the same thing for $-1\lt x\lt0$, so we also get the result for $x\to0^-$
 
1 hour later…
sunny
sunny
12:52
@robjohn Great, so $0<|x|<1$ it is. If you are ambitious, you can add the answer here :) (psst, and you'll receive my upvote!)

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