How can I proof that the divergence of the series $$\sum_{n \in \mathbb N} \frac{|a_n|^2}{1 + \sum_{k=1}^n|a_k|^2}$$ when $\sum_{k \in \mathbb N} |a_n|^2 = \infty$
I've been trying Cauchy test, ratio test and by contradiction, but I couldn't proof yet.
May someone give a hint of why this series diverges?
It's maybe an overkill, but you can prove that the sequence of partial sums of $\sum_{n \in \mathbb N} \frac{|a_n|^2}{1 + \sum_{k=1}^n|a_k|^2}$ is not Cauchy.
Let $S_n=\sum_{k =0}^n |a_k|^2$. Note that $S_n$ is an increasing, nonnegative sequence that goes to $\infty$ as $n$ goes to $\infty$
For any $N\in \mathbb N$ and $q\geq N, \sum_{k=N+1}^q \frac{|a_k|^2}{1 + S_k}\geq \frac{1}{1+S_q} \sum_{k=N+1}^q |a_k|^2=\frac{S_q-S_N}{1+S_q}= 1 - \frac{1+S_N}{1+S_q}$.
For fixed $N$, $\displaystyle \lim_{q\to \infty} \frac{1+S_N}{1+S_q}=0$. You can therefore find some $q\geq N$ such that $\displaystyle \frac{1+S_N}{1+S_q}\leq \frac 12$
For such a $q$, $\displaystyle \sum_{k=N+1}^q \frac{|a_k|^2}{1 + S_k} \geq \frac 12$
Putting everything together, I've proved that $\exists \epsilon > 0, \forall N\in \mathbb N, \exists p\geq N, \exists q\geq N,\left|\displaystyle \sum_{k=p+1}^q \frac{|a_k|^2}{1 + S_k}\right|\geq \epsilon $ (namely, $\epsilon=\frac 12$ and $p=N$)