Can $R[[x]]$ contain constants? 
Consider the ring $R[[x]]$ of formal power series $\sum_{n=0}^\infty a_nx^n$ with coefficients in $R$. I was wondering whether $R[[x]]$ contains elements of $R$ (polynomials of degree $0$). 

I'm trying to solve Commutative Algebra problems. I feel it is possible, as all of $\{a_1,a_2,\dots\}$ simply have to be equal to $0$, which I think is possible.  
Thank you.

If the coefficients are in $A$, we usually write $A[x]$, $A[[x]]$ instead of $R[x]$, $R[[x]]$, and I shall do so here.  This being said:
Yeah, it works.  Formally, $A[[x]]$ is often defined as the set of sequences of elements of $A$; that is, letting $\Bbb Z_0 = \{n \in \Bbb Z \mid n \ge 0\}$, the set of maps from $\Bbb Z_0 \to A$.  The sequences are added pointwise, so that $(a_i) + (b_i) = (a_i + b_i)$, and multiplied according to the rule $(a_i) \times (b_i) = (\sum_{j + k = i} a_j b_k)$, where $i$ is the sequence index in the sum on the right.  $A[x]$ is similarly defined, except the sequences in this case must stabilize to $0$ after some finite index is reached; that is, $(a_i) \in A[x]$ if $(a_i) \in A[[x]]$ and $a_j = 0$ for $j > N \in \Bbb Z_0$.  The least such $N$ is in fact the degree of $(a_i) \in A[x]$.  Soooooo . . . it is easy to see that in either case the ring $A$ (or an isomorphic copy thereof!) may be considered to be the sequences $(a_i)$ with $a_i = 0$ if $i > 0$.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
