Showing a linear operator is compact

Homework Statement

Let $[e_{j}:j\in N]$ be an orthonormal set in a Hilbert space H and $\lambda_{k} \in R$ with $\lambda_{k} \rightarrow 0$

Then show that $Ax=\sum_{j=1}^{\infty} \lambda_{j}(x,e_{j})e_{j}$

Defines a compact self adjoint operator $H \rightarrow H$

The Attempt at a Solution

I've proven that it's self adjoint by extending the orthonormal set to an orthonormal basis and taking an arbitrary y in H and taking the inner product of that y written as a summation using the basis with Ax and then comparing that with the inner product of x with Ay.

To show that it's compact I need to consider some bounded sequence $x_{n}$ and show that $Ax_{n}=\sum_{j=1}^{\infty}\lambda_{j}(x_{n},e_{j})e_{j}$ has some subsequence $Ax_{n_{i}}$ which is convergent.

I'm not sure what direction to take here. Perhaps use the definition that A is compact if the image of the unit ball under A is sequentially compact? Or perhaps taking some the restriction of the sequence $x_n$ to some subset ($Ax_{n}>0$, or if that is empty, $Ax_{n} \leq 0$) and using that $\lambda_{k} \rightarrow 0$ and x_{n} is bounded to show that this subsequence is cauchy and therefore convergent.

I'm not confident with proving that linear operators are compact in general.