Convergence in topology There is a sequence=(1,1,1,1,.....)
                                   x2=(0,1/2,1/2,1/2,...)
                                   x3=(0,0,1/3,1/3,....) and so on.
simply it converges in Product topology.The uniform metric is not bounded so it is not convergent in uniform topology.So I believe it is not convergent in box topology as it is not convergent in uniform topology.But I see that all the open neighbourhood containing (0,0,0,0,.....) contains all the terms except those starting finite terms ie. convergent in box topology.What have I missed? Am I said something which is conceptually wrong!
 A: The basis elements of the uniform topology are sets that look like this:
$U=B(x_1, \epsilon) \times B(x_2, \epsilon) \times B(x_3, \epsilon) \times \ldots$.
In particular the $\epsilon$ value is fixed.
On the other hand, the basis elements of the box topology are sets that look like this:
$V = B(x_1, \epsilon_1) \times B(x_2, \epsilon_2) \times B(x_3, \epsilon_3) \times \ldots$.
Now the $\epsilon$ value varies at each coordinate.
To show $x_n$ non-convergent in the box topology, you need to choose the sequence $\epsilon_1, \epsilon_2, \ldots$ so it decreases quickly enough to make $V$ exclude all of $x_1,x_2,\ldots $ . 
How quickly it needs to decrease I leave that up to you!  
A: $(x_n)_n \rightarrow x, n \to \infty$ in the product topology iff for each coordinate $k$, we have that $(x_n)_k \rightarrow x_k, n \to \infty$.
In "your" sequence (or Munkres' really IIRC) for a fixed coordinate $k$ your sequence is just $0, 0,\dots, 0 $ ($k-1$ many times), ${1 \over n}, {1 \over (n+1)} \ldots$ which indeed tends to $0$.
In the uniform metric we just have to compute $d(x_n, 0)$ where $d$ is the uniform metric. $d(x_n, 0) =\frac{1}{n} \to 0$ as is easily computed so $(x_n)$ does converge to the $0$-sequence in the uniform metric.
The set $O = (-1,1) \times (-\frac{1}{2}, \frac{1}{2}) \times (-\frac{1}{3}, \frac{1}{3}) \ldots = \prod_n U_n$ where $U_n = (-\frac{1}{n}, \frac{1}{n})$ is an open neighbourhood of $0$ in the box topology, that contains no point $x_n$. So no convergence in that topology.
