Convergence of Sequences in $\mathbb{R}^\omega$ In which of the product, uniform, and box topologies do the sequences converge? 
\begin{align*}
a_1 & = (0,1,0,0,0,0,0,0,0,0,...)\\
a_2 & = (0,0,1,2,0,0,0,0,0,0,...)\\
a_3 & = (0,0,0,1,2,3,0,0,0,0,...)\\
a_4 & = (0,0,0,0,1,2,3,4,0,0,...)\\
&.\\
&.\\
&.\\
\end{align*}
where $a_{nk} = k-n$ if $n < k \leq2n$ and $0$ else.
\begin{align*}
b_1 & = (0,1,0,0,0,0,0,0,0,0,...)\\
b_2 & = (0,0,\frac{1}{3},\frac{1}{2},0,0,0,0,0,0,...)\\
b_3 & = (0,0,0,\frac{1}{5},\frac{1}{4},\frac{1}{3},0,0,0,0,...)\\
b_4 & = (0,0,0,0,\frac{1}{7},\frac{1}{6},\frac{1}{5},\frac{1}{4},0,0,...)\\
&.\\
&.\\
&.\\
\end{align*}
where $b_{nk} = \frac{1}{3n-k}$ if $n<k\leq 2n$ $0$ else. 
I confused as to how to prove these sorts of problems. Intuitively I think both are converging in the product topology and I think the second only converges in the uniform, while neither does in the box. I can't seem to figure out how these proofs work.
Thanks for the help!  
 A: Both sequence converge pointwise to the zero sequence $z=\langle 0,0,0,\ldots\rangle$, so that’s the only element of $\Bbb R^\omega$ to which either could converge in any of the three topologies. This is something that you should prove if you’ve not already done so.
You are correct in thinking that both sequences converge to $z$ in the product topology. I’ll prove this for the first sequence as an illustration.

For $x\in\Bbb R^\omega$, $m\in\Bbb Z^+$, and $\epsilon>0$ let $$B(x,m,\epsilon)=\{y\in\Bbb R^\omega:|y_k-x_k|<\epsilon\text{ for all }k\le m\}\;;$$ the family of all such sets forms a base for the product topology on $\Bbb R^\omega$. (Why?) Let $U$ be an arbitrary open nbhd of $z$; we want to show that there is an $m\in\Bbb Z^+$ such that $a_n\in U$ for each $n\ge m$.
$U$ is an open nbhd of $z$, so there are an $m\in\Bbb Z^+$ and an $\epsilon>0$ such that $B(z,m,\epsilon)\subseteq U$. Suppose that $n\ge m$. If $k\le m$, then $k\le n$, so by definition $a_{nk}=0$, and $$|a_{nk}-z_k|=|0-0|=0<\epsilon\;,$$ so $a_n\in B(z,m,\epsilon)\subseteq U$. Thus, $\langle a_n:n\in\Bbb Z^+\rangle$ does indeed converge to $z$ in the product topology.

You can use a similar argument to show that $\langle b_n:n\in\Bbb Z^+\rangle\to z$ in the product topology.
You are also correct in thinking that only the second converges in the uniform topology. For $x\in\Bbb R^\omega$ and $\epsilon>0$ let $$B(x,\epsilon)=\{y\in\Bbb R^\omega:|y_n-x_n|<\epsilon\text{ for all }n\in\Bbb Z^+\}\;;$$ the family of all such sets is a base for the uniform topology on $\Bbb R^\omega$. (Why?) Now just observe that not only does the sequence $\langle a_n:n\in\Bbb Z^+\rangle$ not lie in $B(z,1)$ from some point on, $B(z,1)$ actually contains no point of the sequence. Since $B(z,1)$ is an open nbhd of $z$, this certainly shows that $\langle a_n:n\in\Bbb Z^+\rangle$ does not converge to $z$ and hence (by the remark at the beginning) does not converge at all in the uniform topology.
To show that $\langle b_n:n\in\Bbb Z^+\rangle\to z$ in the uniform topology, use the fact that $0\le b_{nk}\le\frac1n$ for each $n,k\in\Bbb Z^+$ to show that every uniform open nbhd of $z$ contains a tail of the sequence; the basic open sets $B\left(z,\frac1n\right)$ should prove useful.
Finally, you are correct in thinking that neither converges in the box topology. Let
$$U=\prod_{n\ge 1}\left(-\frac1{2n},\frac1{2n}\right)=\left(-\frac12,\frac12\right)\times\left(-\frac14,\frac14\right)\times\left(-\frac16,\frac16\right)\times\ldots\;.$$
Verify that $U$ is an open nbhd of $z$ in the box topology on $\Bbb R^\omega$, and use this to show that neither sequence converges to $z$.
