Show if $\lim_{n \to \infty} \lambda_n=0$ then $Tu=\sum^\infty_{n=1} \lambda_n \langle u,e_n \rangle e_n$ defines a compact operator. Let $\{e_n\}$ be an orthonormal basis in a Hilbert space $H$ and let $\{\lambda_n\}$ be a sequence of numbers. 
Define the operator $$T:H \to H$$ by $$Tu=\sum^\infty_{n=1} \lambda_n \langle u,e_n \rangle e_n$$ where $u \in H$
I am trying to show that if $\lim_{n \to \infty} \lambda_n=0$ then $T$ is a compact operator. 

Why does showing $$||Tu-T_k u||^2 \leq \sup_{n >k} \{|\lambda_n|^2\}
 ||u||^2$$ tell us that $|T-T_k u|| \to 0$?

Where $T_k$ is $k^{th}$ partial sum in the definition of $T$.

I have the full proof for this but I am wondering why does showing 
  $\lim_{n \to \infty} \lambda_n=0$ mean that $T$ is a compact operator?
How do you show that a operator is compact?

I have come across that statement  that a compact operator is the limit of finite dimensional operators but I do not understand this statement. 

What does this statement mean?

 A: Let $X,Y$ be Hilbert spaces and let $T:X \to Y$ be a bounded linear operator, then we say that $T$ is compact if one, and hence all, of the following conditions hold:


*

*$T$ is the norm limit of a sequence $(T_n)$ of finite rank operators;

*For every bounded subset $V$ of $X$, $TV$ is relatively compact in $Y$;

*For each bounded sequence $(x_n)$ in $X$, there is a subsequence of $(Tx_n)$ which converges in $Y$.


In example, the first criterion is used: Let $T:H \to H$ be defined by 
$$Tu = \sum_{n=1}^\infty\lambda_n(u,e_n)e_n, \quad u \in H$$
and define $T_k :H \to H$ by
$$T_ku = \sum_{n=1}^k\lambda_n (u,e_n)e_n$$
We will show that 
$$\|T - T_k\| \to 0, \quad n \to \infty$$
To do so, let $u \in H$, then you already have that
$$\|Tu - T_ku\|^2 \leq \sup_{n \gt k} \{ |\lambda_n|^2\}\|u\|^2$$
which says, by definition, that
$$\|T-T_k\|^2 \leq \sup_{n \gt k} \{ |\lambda_n|^2\}$$
Since $\lambda_n \to 0$, we have that
$$\sup_{n \gt k} \{ |\lambda_n|^2\} \to 0, \quad k \to \infty$$
In other words,
$$\|T-T_k\|^2 \to 0, \quad k \to \infty$$
And hence we conclude that $T$ is compact.
