Topology defined by some convergent sequences On $\mathbb R^d$ define this notion of convergence
$$x_n\to x\iff \lim_n ||x_n-x||_2 n^{d}=0$$
is there any topology which is endowed with this convergence? 
It has the peculiarity that if $x_n$ converges to $x$ in the standard sense you can find a subsequence of $x_n$ such that $x_n\to x$, so balls cannot be open sets
 A: Edit: As pointed out by @Kavi Rama Murthy, there is a serious flaw in this argument. I'll have to get back to it to see if I'm even right in that it doesn't come from a topology.
There is not.  Take the sequence $n^{-(d+1)}$, which converges to 0 in your definition.  
Now look at the sequence $(y_n) = (1,1,2^{-(d+1)}, 2^{-(d+1)}, 3^{-(d+1)}, 3^{-(d+1)},...)$. If your convergence comes from a topology, this sequences would also converge to 0. But 
$||y_n - 0||n^d \ge \left(\frac{n-1}{2}\right)^{-(d+1)}n^d=2^{d+1}(n-1)^{-(d+1)}n^d=2^{d+1}\frac{1}{n-1}\left(\frac{n}{n-1}\right)^d$
A: This solution steals the idea from Richard Jensen's answer, but then (hopefully!) correctly proves the conclusion that the convergence notion cannot come from a topology.
Define for $n \ge 1:$ $p_n:=n^{-d-1}$ and $x_n:=(p_n,0,\ldots,0) \in \mathbb R^d$. Then $x_n \rightarrow (0,\ldots,0) \in \mathbb R^d$ in the sense of the problem's definition of convergence.
Now define $y_n:=x_{\lfloor \sqrt{n} \rfloor}$. Since $\lim_{n \to \infty}\lfloor \sqrt{n} \rfloor = \infty$, the sequence $\{y_n\}$ must also converge to $(0,\ldots,0) \in \mathbb R^d$, if this convergence notion is coming from a topology.
But we have for $k\ge 1$
$$\lVert y_{k^2}-(0,\ldots,0)\rVert_2(k^2)^d = \lVert x_k-(0,\ldots,0)\rVert_2(k^2)^d =p_k(k^2)^d =k^{-d-1}k^{2d}=k^{d-1} \ge 1.$$
So the sequence $\{y_n\}$ is clearly not converging to $(0,\ldots,0) \in \mathbb R^d$, which is a contradiction.
