Suppose one has infnite and countable $\{b_j\in \mathbb{R}:j=1,2,\ldots,\}$ real numbers. Furthermore, one has $$ \sum_{j=1}^{\infty} b_j^2 <\infty $$ Can we say that $$ \sum_{j=1}^{\infty} b_j <\infty $$ is true?
My thought so far is this:
By Cauchy's criterion, for any $\varepsilon>0$, there exists a $K$ such that
$$ \sum_{j=m+1}^n b_j^2 < \varepsilon $$
whenever $n>m>K$. This means, we could find a minimum value $w$ of $\{b_j^2>0: j =m+1,\ldots,n\}$ such that
$$ \sqrt{w}\sum_{j=m+1}^n |b_j| < \sum_{j=m+1}^n b_j^2 < \varepsilon $$
Thus,
$$ \sum_{j=m+1}^n b_j< \sum_{j=m+1}^n |b_j|< \varepsilon^{'}=\varepsilon/\sqrt{w}$$
Since $\varepsilon$ is arbitrary, so does $\varepsilon^{'}$.
This, using Cauchy's criterion again, implies that
$$ \sum_{j=1}^{\infty} b_j < \sum_{j=1}^{\infty} |b_j| <\infty$$
Short answer is that the statement is false.
The mistake in your proof was in the second last line 'Since $\varepsilon$ is arbitrary, so does $\varepsilon '$'. In this case, $\varepsilon'$ is not arbitrary, since it depends on $w$ which depends on $K$. Strictly speaking, say I pick $\eta >0$, you'd wanna pick $K>0$ so that $\forall n>m>K$, $$\left|\sum_{j=m+1}^n b_j \right|<\eta$$ What $K$ will you pick? If you use your previous argument, you'd want to pick $K$ so that $\forall n>m>K$, $$\sum_{j=m+1}^n b_j^2<\eta \sqrt{w} \qquad (1)$$ which gives $$\left|\sum_{j=m+1}^n b_j \right|\le \frac{\sqrt{w}}{\sqrt{w}}\sum_{j=m+1}^n \left|b_j \right|\le \frac{1}{\sqrt{w}}\sum_{j=m+1}^n b_j^2<\eta\sqrt{w}/\sqrt{w} =\eta$$ But you may not have been able to pick such a $K$ for equation $(1)$ to hold. Depending on what $K$ you pick, the $\sqrt{w}$ changes. This is where the argument breaks down.
Intuitively, squaring the terms does not necessarily make the sum larger. That's why the statement does not hold.