Let $X$ be a Banach space and suppose we have a sequence $\{x_n\}$ which is convergent weakly but not strongly. Define $y_n:=\sum\limits_{k=1}^{n}x_k$.
What we can say about $\limsup\limits_{n\to\infty} \|y_n\|^{\frac{1}{n}}$ and $\liminf\limits_{n\to\infty}\|y_n\|^{\frac{1}{n}}$ ?
Sequence $x_n$ is bounded, so $M := \limsup \|x_n\|$ exists.
$0 \le \limsup \|y_n\| \le \infty$ with either equality possible. Same for $\liminf \|y_n\|$.
Let $z_n = \frac{1}{n}\sum_{k=1}^n x_n = \frac{1}{n} y_n$. Then $0 \le \limsup \|z_n\| \le M$ with either equality possible. Same for $\liminf \|z_n\|$.
Weakly convergent sequences are bounded, say $\lVert x_n\rVert \leqslant B$ for all $n$. Since we assume that $(x_n)$ is not norm-convergent, it follows that $B > 0$. Then
$$\lVert y_n\rVert \leqslant \sum_{k = 1}^n \lVert x_k\rVert \leqslant n\cdot B,$$
whence
$$\limsup_{n\to\infty}\, \lVert y_n\rVert^{\frac{1}{n}} \leqslant \limsup_{n\to\infty} (n\cdot B)^{\frac{1}{n}} = 1.\tag{1}$$
If we had $\limsup\limits_{n\to\infty}\, \lVert y_n\rVert^{\frac{1}{n}} = c < 1$, then for all large enough $n$ we'd have $\lVert y_n\rVert^{\frac{1}{n}} < d := \frac{1+c}{2} < 1$, and so $\lVert y_n\rVert < d^n \to 0$, which implies $y_n \to 0$ in norm, and hence also $x_n = y_n - y_{n-1} \to 0$ in norm. Thus for a sequence $(x_n)$ that is weakly convergent but not norm-convergent, we always have
$$\limsup_{n\to\infty} \, \lVert y_n\rVert^{\frac{1}{n}} = 1\tag{2}$$
with $y_n = \sum\limits_{k = 1}^n x_k$.
Clearly we have
$$0 \leqslant \liminf_{n\to \infty}\, \lVert y_n\rVert^{\frac{1}{n}} \leqslant \limsup_{n\to \infty}\, \lVert y_n\rVert^{\frac{1}{n}} = 1\tag{3}$$
in our situation. It is easy to construct examples where the extrema are attained. For example in $\ell^2(\mathbb{N})$ we can take $x_n = e_n$, so $x_n \rightharpoonup 0$, and $\lVert y_n\rVert = \sqrt{n}$ shows $\liminf\limits_{n\to \infty}\,\lVert y_n\rVert^{\frac{1}{n}} = 1$. And choosing $x_n = (-1)^{n+1}\cdot e_{\lceil n/2\rceil}$ gives us $y_{2m} = 0$ for all $m$, and so $\liminf\limits_{n\to \infty} \, \lVert y_n\rVert^{\frac{1}{n}} = 0$. A similar construction gives us a sequence converging to $0$ weakly but not in norm with $\liminf\limits_{n\to\infty}\, \lVert y_n\rVert^{\frac{1}{n}} = c$ for every $c \in (0,1)$: $x_1 = e_1$, $x_{2m} = c^{2m}e_{m+1} - e_m$ and $x_{2m+1} = (1-c^{2m})e_{m+1}$ gives us a sequence with $y_{2m} = c^{2m}e_{m+1}$ and $y_{2m-1} = e_m$. Thus we cannot say more than $(3)$ about $\liminf\limits_{n\to \infty} \, \lVert y_n\rVert^{\frac{1}{n}}$ without knowing the sequence $(x_n)$.
