= Solution
A <compact operator> maps every bounded sequence to a sequence with a norm-convergent subsequence. A <Hilbertian basis> is a complete orthonormal sequence.
If $T$ is compact and $\lambda_n$ does not tend to zero, some subsequence has
$|\lambda_{n_j}|\geq\varepsilon$. The vectors
$Te_{n_j}=\lambda_{n_j}e_{n_j}$ are pairwise separated by at least $\sqrt2\varepsilon$, so they have no convergent subsequence, a contradiction.
Conversely, if $\lambda_n\to0$, define the <finite-rank operator>
$$
T_Ne_n=\begin{cases}\lambda_ne_n,&n\leq N,\\0,&n>N.\end{cases}
$$
Then
$$
\|T-T_N\|=\sup_{n>N}|\lambda_n|\longrightarrow0.
$$
An operator-norm limit of compact operators is compact, so $T$ is compact.
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