An arctan criterion for convergence? Is the following inference correct, and if so, is it a mere curiosity?
Let $\{a_k\}_{k \in \Bbb N}$ be a sequence of positive real numbers and set
$$
P_k = \sum_{j=0}^{k-1}  a_k \quad \mbox{and} \quad
Q_k = \sum_{j=0}^{k-1}  a_k^2.
$$
Then the partial sums $P_k$ converge if and only if the series:
$$
\sum_{n=0}^{\infty} \arctan \frac{a_n}{\sqrt{Q_n}}
$$
converges.
 A: Suppose first that the series $\displaystyle \sum a_k$ is convergent. Then $a_n\to 0$, and we have $\sqrt{Q_n}\geq a_0>0$ for $n\geq 1$. Hence $\displaystyle 0< b_n=\frac{a_n}{\sqrt{Q_n}}\leq \frac{a_n}{a_0}$, and $b_n\to 0$. If $f(x)={\rm Arctan}(x)$, we have if $x\to 0$ that $f(x)\sim x$; hence $c_n=f(b_n)\sim b_n$. As  $\displaystyle \frac{a_n}{a_0}$ is a convergent series, this imply that $b_n\leq \frac{a_n}{a_0}$ also, and as all the terms are positive, that the series $c_n$ is convergent.
Suppose now that the series $\displaystyle c_n={\rm Arctan}(\frac{a_n}{\sqrt{Q_n}})$ is convergent. Then $c_n\to 0$, hence $\displaystyle \frac{a_n}{\sqrt{Q_n}}\to 0$, and we have $\displaystyle c_n \sim \frac{a_n}{\sqrt{Q_n}}=b_n$. This imply that the series $\sum b_n$ is convergent, and also is the series $\sum b_n^2$.
Now we have $ Q_{n+1}-Q_n=a_n^2$. Hence:
$$ b_n^2=\frac{a_n^2}{Q_n}=\frac{Q_{n+1}-Q_n}{Q_n}=\frac{Q_{n+1}}{Q_n}-1$$
Hence $\displaystyle \frac{Q_{n+1}}{Q_n}=1+b_n^2$, and we get for $m\geq 1$
$$\frac{Q_{m+1}}{Q_1}=\prod_{k=1}^m (1+b_k^2)$$
But the series $\displaystyle \sum \log(1+b_k^2)$ is convergent (it is positive and $\sim b_n^2$). Thus we get that $Q_n$ is a convergent sequence, say to $L>0$. Hence $a_n=\sqrt{Q_n}b_n\sim \sqrt{L}b_n$, and the series $\sum a_n$ is convergent.
