First, the authors give a refinement of the result due to \textit{R.\,Aron, D.\,García} and \textit{M.\,Maestre} [J.~Convex Anal.\ 16, 667--672 (2009; Zbl 1191.46014)] that in every separable Banach space \(X\) there is a weakly dense sequence that tends to infinity in norm. Namely, they show that, if for a sequence \((a_n)\) of positive reals which tends to infinity there is a sequence \((x_n)\subset X\), \(\|x_n\|=a_n\), with a weak cluster point, then there is a weakly dense sequence \((y_n)\subset X\) with \(\|y_n\|=a_n\). After that, the authors consider the question: If a sequence has a weak limit with respect to a given filter \(\mathcal{F}\), how quickly can the norms of the elements in the sequence tend to infinity? It is known that the answer depends on the filter. A sequence in a Hilbert space which converges with respect to the filter of weak statistical convergence can tend to infinity in norm [\textit{J.\,Connor, M.\,Ganichev} and \textit{V.\,Kadets}, J.~Math.\ Anal.\ Appl., 244, 251--261 (2000; Zbl 0982.46007)]. On the other hand, for some filters a weakly \(\mathcal{F}\)-convergent sequence cannot go to infinity in norm [\textit{M.\,Ganichev} and \textit{V.\,Kadets}, ``Filter convergence in Banach spaces and generalized bases'', in: T.\,Banakh (ed.), ``General Topology in Banach Spaces'', Nova Sci.\ Publ.\, Huntington, NY, 61--69 (2001; Zbl 1035.46009)]. The authors give a very general answer to the mentioned question. Then they apply this result and show, in particular, that for every weakly statistically convergent sequence \((x_n)\) with increasing norms in a Hilbert space, we have \(\sup_n\|x_n\|/\sqrt{n}<\infty\), and that this estimate is sharp. They also consider other types of weak filter convergence, in particular, the Erdős-Ulam filters, analytical \(P\)-filters and \(F_{\sigma}\) filters.