Let $A$ be a compact, self-adjoint operator on a Hilbert space $H$. Then, the spectrum theorem says that for every $x\in H$
$$Ax = \sum_{k=1}^{\infty} \lambda_k\langle x,e_k \rangle e_k$$ where $\lambda_k$ are nonzero eigenvalues and $e_k$ are the corresponding orthnormal eigenvectors.
Then, the book that I found ("Applied Analysis" by Hunter) says that assuming $A$ has an infinite sequence of nonzero eigenvalues,
(1) the range of $A$ is
$$\text{ran A} = \big\{ \sum_{k=1}^{\infty} c_ke_k \mid \sum_{k=1}^{\infty} \frac{|c_k|^2}{|\lambda_k|^2} < \infty \}$$
and
(2) the range is not closed since $\lambda_n \to 0$ as $n \to \infty$
But, I do not understand (1) and (2).
For (1), how do I get the condition $\sum_{k=1}^{\infty} \frac{|c_k|^2}{|\lambda_k|^2} < \infty $?
For (2), how does the fact that $\lambda_n \to 0$ as $n \to \infty$ show that the range is not closed?
Any element $Ax = \sum_{k=1}^{\infty} c_k e_k$ of the range has $c_k = \lambda_k \langle x, e_k \rangle$ and therefore $$\sum_{k=1}^{\infty} \frac{|c_k|^2}{|\lambda_k|^2} = \sum_{k=1}^{\infty} |\langle x, e_k \rangle |^2 \le \langle x,x \rangle < \infty$$ by Bessel's inequality. Conversely, if $\sum_{k=1}^{\infty} \frac{|c_k|^2}{|\lambda_k|^2}$ converges then you can let $x = \sum \frac{c_k}{\lambda_k} e_k.$
The second part holds more generally: the range of any compact operator, if it's infinite-dimensional, is not closed, which follows from the open mapping theorem since the closure of an open subset is not compact unless the space is finite-dimensional.