Graded tensor algebra Given a finite dimensional $\mathbb R$-vectorspace $V$ we can make 
$$
T(V) := \bigoplus_{n=0}^\infty V^{\otimes n}.
$$ Here $V^{\otimes n} =  V \otimes \cdots \otimes V$. An element of $T(V)$ looks like $(v_i)_i$ where $v_i \in V^{\otimes i}$ and $v_i \neq 0$ for only finitely many $i$. I don't understand how $T(V)$ is made into a graded algebra, in particular how the multiplication $\cdot : T(V) \times T(V) \to T(V)$ is given, i.e., what is $(v_i)_i \cdot (w_i)_i$ ?

If we write $(v_i)_i$ and $(w_i)_i$ as 
$$
(v_i)_i = \sum v_i e_i, \qquad (w_i)_i = \sum w_i e_i
$$ can we define 
$$
 (v_i)\cdot(w_i) := \sum_{i=0}^\infty \left ( \sum_{k=0}^i v_kw_{i-k} \right ) e_i?
$$
 A: You concatenate the terms. If $v=v_1\otimes\cdots\otimes v_m$ and $w=w_1\otimes\cdots\otimes w_n$ are some pure elements of $T^mV=V^{\otimes m}$ and $T^nV=V^{\otimes n}$ respectively, then their product is
$$w\times w=v_1\otimes\cdots\otimes v_m\otimes w_1\otimes\cdots\otimes w_n\in T^{m+n}V$$
The product is then extended to the whole of $TV$ by bilinearity (not all elements of $T^iV=V^{\otimes i}$ are pure, but they are linear combinations of pure elements). This product is well defined.
EDIT. Let's take $v,w\in TV$. Since $TV=\bigoplus_{n=0}^{\infty}V^{\otimes n}$, we get that there exists an integer $N$ such that $v,w\in\bigoplus_{n=0}^{N}V^{\otimes n}$. Thus there are $v_0,w_0\in\Bbb R$, $v_1,w_1\in V$, $v_2,w_2\in V\otimes V,\dots$,$v_N,w_N\in V^{\otimes N}$ such that $v=\sum_{n=0}^{N}v_n$ and $w=\sum_{n=0}^{N}w_n$.
Let $x$ be either the letter $v$ or the letter $w$. Then for any $0\leq n\leq N$ there exists pure tensors $x_{n,i}$, $i=1,\dots, M$ (for some large $M$ that will fit everyone) such that $x_n=\sum_{i=0}^Mx_{n,i}$.
To extend the product by bilinearity means that we will define $v\times w$ for arbitrary elements of the tensor algebra by expanding the product through distributivity until we obtain a sum of products of pure tensors (which are already defined). It then still remains to prove that the result is independent of the (non unique) decomposition into pure tensors chosen for $v$ and $w$. Here this will translate to
$$\begin{array}{rcl}
v\times w&\stackrel{\text{def}}{=}&\left(\sum_{n=0}^{N}v_n\right) \times \left(\sum_{n=0}^{N}w_n\right)\\
&=&\sum_{k,l=0}^Nv_k\times w_l\\
\end{array}$$
where for any $k,l\in\lbrace 0,\dots,N\rbrace$
$$\begin{array}{rcl}
v_k\times w_l&\stackrel{\text{def}}{=}&\left(\sum_{i=0}^{M}v_{k,i}\right) \times \left(\sum_{j=0}^{N}w_{l,j}\right)\\
&=&\sum_{i,j=0}^M v_{k,i}\times w_{l,j}\\
\end{array}$$
Here every term is already defined, since the $v_{k,i},w_{l,j}$ are by definition pure tensors (of degree $k$ and $l$ respectively), and we have defined the product of two pure tensors above.

So for instance, if $V$ is three dimensional with basis $a,b,c$, then below is an example of a product of non pure tensors being carried out
$$\begin{array}{rl}
\big(a\otimes a+7c\otimes b\otimes a\big)
\times
\big(2a+b+b\otimes b\otimes c\big)&=&\\
a\otimes a\otimes(2a+b)+a\otimes a\otimes b\otimes b\otimes c\\
+7c\otimes b\otimes a\otimes(2a+b) +7c\otimes b\otimes a\otimes b\otimes b\otimes c
\end{array}$$
A: The canonical definition (unlike the one given in other answer) is that it is the bilinear map associated to the canonical natural functorial tensorial isomorphism given by associativity.
