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Is there a locally convex space $(X, \tau)$ for which there is a weakly convergent sequence $x_n$ which is not contained in the closed absolutely convex hull of a $\tau$-convergent sequence $y_n$?

Clearly, we can restrict the search to Mackey spaces since the Mackey topology $\mu(X,X') \supseteq \tau$ has less convergent sequences than $\tau$. On the other hand, such a space $(X, \tau)$ must be necessarily non-metrizable since in a metrizable space for every weakly convergent sequence $x_n$ there is a $\tau$-convergent sequence $y_n \in conv(x_1, \dots, x_n)$ and thus $x_n$ is contained in the closed convex hull of all the $y_n$. [EDIT: This is obviously wrong.] Also non-metrizable topologies like the finest locally convex topology $\tau = \mu(X,X^*)$ (where $X^*$ is the algebraic dual) is not a counter-example since $\sigma(X,X^*)$-convergent sequences are $\mu(X,X^*)$-convergent.

So we need a non-metrizable Mackey space $X$ with a rather large topology (having only a few convergent sequences), e.g. satisfying (1) $X$ contains an infinite-dimensional weakly null sequence and (2) in $X$ every convergent sequence is finite-dimensional (in which case its closed absolutely convex hull is finite-dimensional). Does there exist such a space?

EDIT: Thanks to Daniel Fischer and Jochen, the answer to the question becomes rather simple.

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    How do you get that $x_n$ belongs to the closed convex hull of all the $y_n$ from $y_n \in \operatorname{conv} (x_1,\dotsc,x_n)$? – 2017-02-07
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    Ohhh, I see, mixed up the sequences: $y_n$ is of course contained in $\overline{\textrm{conv}}\{ x_n \mid n \in \mathbb{N} \}$. Bad mistake. – 2017-02-07
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    In Banach or Frechet spaces, the closed convex hull of a convergent sequence is compact. It is thus enough to have a weakly convergent sequence without $\tau$-convergent subsequences like, e.g., the standard orthonormal basis $(e_n)_{n\in\mathbb N}$ of the Hilbert space $\ell^2$. – 2017-02-07

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