Answer by fedja for Bounding the decrease after applying a contraction operator $n$ vs $n+1$ times

You can easily get a slightly cruder bound $d/n$ (or, if you want, $r/n$) as follows.

Let $A_n=(T^*)^nT^n$ and $B_n=A_n-A_{n+1}$. Then $(B_nv,v)=\|T^nv\|^2-\|T^{n+1}v\|^2\ge 0$, so $B_n$ is positive definite. Also $B_{n+1}=T^*B_nT$, so, since $T$ is a contraction, $Tr B_{n+1}\le Tr B_n$ (this is obvious if $T$ is diagonal but in general $T=R_1DR_2$ where $R_j$ are orthogonal and $D$ is a diagonal contraction and conjugation by an orthogonal matrix doesn't change either trace, or positive definiteness).

So, $n\, Tr B_n\le \sum_{k=1}^n Tr B_k=Tr A_1-Tr A_{n+1}\le Tr A_1\le r$ and we are done because the trace dominates the norm for positive definite matrices.