The question goes as follows:
$T$ has a finite rank $\iff$ $\exists N \in \mathbb{N}$ such that $\lambda_n=0$, $\forall n \geq N$.
Given is the data: $X$ is a Hilbert space with an orthonormal basis $(e_n)$. The linear operator $T: X \rightarrow X$ is defined as; $Tx=\sum\limits_{n=1}^{\infty} \lambda_n(x,e_n)e_n$.
In a previous question I solved $T \in K(X)$ $\iff$ $\lambda_n \rightarrow 0$, which I`m supposed to use.
My attempt:
If $T$ is a compact operator with infinite rank, then $\lambda_n \rightarrow 0$ as $n \rightarrow \infty$. So if $T$ has rank $K$, with $K$ finite, then $\lambda_n \rightarrow 0$ still has to hold. But because no values of $\lambda_n$ are plugged in for $n > K$, thus for $N=K-1$ we have, $\lambda_n=0$ for all $N \in \mathbb{N}$ whenever $n \geq N$.
The $\leftarrow$ way I understand.
By definition, rank of $T$ = dimension of the range $R(T)$. We have $Te_n=\lambda_n e_n$, and if $\lambda_n\ne0$, then $e_n=T(\lambda_n^{-1}e_n)$; i.e., $e_n\in R(T)$. Since $\operatorname{dim} R(T)<\infty$, it follows that $\lambda_n\ne0$ only for finitely many $n$. Thus, $\exists N\colon \lambda_n=0$ for $n>N$. QED.