I think that this operator might not be bounded.
Let $e_i$ denote the i-th vector in the standard orthonormal basis form $\ell^2(\mathbb{N})$. Then
$$ C\cdot e_i=\alpha_i\sum_{j=1}^i e_j $$
thus $||C||\ge\sup_{n}{(\sqrt{n}\alpha_n)}$. However, the rightmost amount might be infinite even if you assume that $\alpha_n$ are positive and that the sum $\sum \alpha_n$ converges.
Indeed, take the sequence defined by $\alpha_n = \frac{1}{log_2(n)^2}$ if $n$ is a power of 2 and $\alpha = 0$ otherwise. The sum $\sum \alpha_n$ becomes $\sum_{k=1}^\infty \frac{1}{k^2}$, which does converge. However, it's clear that
$$ \sup_{n}{(\sqrt{n}\alpha_n)} = \sup_{k} (2^{k/2}*\frac{1}{k^2})=\infty $$
of you want $\alpha_n$ to be strictly positive, then replace $\alpha_n=0$ by $\alpha_n=\frac{1}{2^n}$ (or any rapidly decaying sequence).