Suppose that $T\geq 0$ is a self-adjoint compact operator with norm $1$, then $\lim_{n \rightarrow \infty}T^nx$ exists for all $x \in H$.
Sure with spectral theory for compact operators we obtain the knowledge that $Tg_n = \lambda_n g_n$ for all $n$ where $g_n$ is orthonormal basis of the $H$. By induction we can see that $T^k g_n = \lambda_n^k g_n$.Then for any $x \in H$ we can write $Tx = \sum_{k=1}^\infty \lambda_k\langle x,g_k\rangle g_k$.
Iterate the summation form of the $Tx$ so we obtain $T^2x = \sum_{k=1}^\infty \lambda_k T(\langle x,g_k\rangle g_k) = \sum_{k=1}^\infty \lambda_k^2\langle x,g_k\rangle g_k$, general case follows easily with induction $T^n x = \sum_{k=1}^\infty \lambda_k^n\langle x,g_k\rangle g_k$.
Since a convergent sequence is bounded we have a majorant $\sum_{k=1}^\infty\lambda_k^n \langle x,g_k\rangle g_k \leq \sum_{k=1}^\infty\sup_{k\geq 1}|\lambda_k^n |\langle x,g_k\rangle g_k = \sup_{k\geq 1}|\lambda_k^n | x $. We can change order of summation and limes, namely $\lim_{n \rightarrow \infty}\sum_{k=1}^\infty \lambda_k^n \langle x,g_k\rangle g_k = \sum_{k=1}^\infty \lim_{n \rightarrow \infty}\lambda_k^n \langle x,g_k\rangle g_k$, but $\lim_{n \rightarrow \infty}\lambda_k^n$ does not converge if $|\lambda_k| > 1$ for some $k$.
There must be something wrong with my argument since I didn't use the fact that $\|T\| = 1$ and $\lim_{n \rightarrow \infty} T^nx = 0$ for all $x \in H$ would be quite boring result.
It is good to start off with some observations. As $T$ is a self-adjoint compact operator on a separable infinite-dimensional Hilbert space $H$, there exists an orthonormal basis $\{e_{k}:k\in\mathbb{N} \}$ where distinct indices correspond to distinct vectors, such that the orthonormal basis consists of eigenvectors of $T$. For each $k\in\mathbb{N}$, let $\lambda_{k}$ be the eigenvalue corresponding to the eigenvector $e_{k}$.
In addition, note that the that all of the eigenvalues are contained in the interval $[0,1]$. The eigenvalues are contained in $[0,\infty )$ because $T$ is a positive operator, and they are further contained in $B_{H}$ because $T$ has norm one. So it follows that the eigenvalues of $T$ are contained in $[0, \infty )\cap B_{H} = [0,1]$.
If $k,n\in\mathbb{N}$, then $e_{k}$ is an eigenvector of $T^{n}$ with eigenvalue $\lambda_{k}^{n}$. It follows that as $\lambda_{k}\in [0,1]$ for each $k\in\mathbb{N}$, $\lim_{n\to\infty}\lambda_{k}^{n}$ exists for every $k\in\mathbb{N}$. Let $\mu_{k}$ denote this limit.
Furthermore, as $T$ is a compact operator, the set $\{n\in\mathbb{N}: \lambda_{n} > 1/2 \}$ is finite. So it follows that there is some $C\in (0,1)$ such that all of the eigenvalues contained in $[0,1)$ satisfy $\lambda \leq C$. From this, it can be shown that given $\varepsilon >0$, there exists some $N\in\mathbb{N}$ such that if $n\geq N$ and $k\in\mathbb{N}$, then $|\lambda_{k}^{n} - \mu_{k}| \leq \varepsilon$.
Now define $S:H\to H$ by $S(x) := \sum_{n=1}^{\infty}\mu_{k}\langle x, e_{k}\rangle e_{k}$. As $\{k\in\mathbb{N} : \mu_{k} \neq 0\}$ is finite, it follows that $S$ is well-defined and is linear. In fact, $S$ is a projection onto a finite-dimensional subspace of $H$, because if $\mu_{k}\neq 0$ for some $k\in\mathbb{N}$, then $\mu_{k} = 1$.
It will now be shown that $\lim_{n\to\infty}T^{n}(x) = S(x)$ for every $x\in H$. To show this, let $x\in H$. Let $\varepsilon >0$. As shown above, there exists some $N\in\mathbb{N}$ such that $n\geq N$ and $k\in\mathbb{N}$ implies $|\lambda_{k}^{n} - \mu_{k}| \leq \varepsilon$. It follows that if $n\geq N$,
\begin{align*} \|T^{n}(x)-S(x)\|^{2} &= \left\| \sum_{k=1}^{\infty} (\lambda_{k}^{n} - \mu_{k})\langle x, e_{k}\rangle e_{k} \right\|^{2} \\ &= \sum_{k=1}^{\infty} \|(\lambda_{k}^{n} - \mu_{k})\langle x, e_{k}\rangle e_{k}\|^{2} \\ &= \sum_{k=1}^{\infty}|\lambda_{k}^{n} - \mu_{k}|^{2} |\langle x, e_{k}\rangle |^{2} \\ &\leq \varepsilon^{2}\sum_{k=1}^{\infty} |\langle x, e_{k}\rangle |^{2} \\ &= \varepsilon^{2}\|x\|^{2}. \end{align*}
Consequently, $\|T^{n}(x) - S(x)\| \leq \varepsilon \|x\|$ for all $n\in\mathbb{N}$ with $n\geq N$. This shows that $\lim_{n\to\infty}T^{n}(x) = S(x)$, so that as $x$ was arbitrary, $\lim_{n\to\infty}T^{n}(x)$ exists for each $x\in H$. In fact, the sequence $(T^{n})_{n\in\mathbb{N}}$ converges pointwise to $S$. Even further, the convergence is actually in norm by a corollary of the uniform boundedness theorem, although that can also be shown directly from an inequality obtained above.
In particular, you have the even stronger result that the sequence $(T^{n})_{n\in\mathbb{N}}$ converges in norm to a bounded operator which projects onto a finite-dimensional subspace.