Paradox of summation identities and infinity? First: someone can proof that square of sum is:
$(a_1+a_2+...+a_n)^2=(a^2_1+a^2_2+\cdots+a^2_n)+2(a_1a_2+...+a_{n-1}a_{n})$
in other words:
$$(\sum_{i=1}^na_i)^2=\sum_{i=1}^na^2_i+2\sum_{i}^{n-1}\sum_{j=i+1}^na_ia_j$$
Second: what happens to this term of summation $2\sum_{i=1}^{n-1}\sum_{j=i+1}^na_ia_j$ if n goes to infinity $n\to\infty$, that which means that $n-1=n$ and it seems to me that the last term of summation collapse, can someone tell me what happens to last term?
 A: The first part is a simple induction where the base is Newton's binomial expansion and the step is just:
$$\begin{align*} (a_1+\cdots+a_n) ^2 
=& ((a_1+\cdots+a_{n-1})+a_n) ^2\\ 
=& (a_1+\cdots+a_{n-1})^2 + 2(a_1+\cdots+a_{n-1})a_n + a_n^2
\end{align*}$$ 
(just substitute $(a_1,\cdots,a_{n-1})^2$ by the formula you gave and reorganize the terms)
As for the second part, for the whole thing to be well defined when you take the limit the series $\sum_{n=1}^\infty a_n$ has to converge (which guarantees convergence of the first term) as for the second term, you can rearrange it as 
$$ 2 \sum_{i=1}^\infty a_i (\sum_{j=i+1}^\infty a_j)$$
Since the series converges $\sum_{j=i+1}^\infty a_j\to 0$ with $i\to \infty$ in paricular, for all $i$ bigger then some $N$ $|\sum_{j=i+1}^\infty a_j|<1$ and thus we can separate it into:
$$\begin{align*} 2 \sum_{i=1}^\infty a_i (\sum_{j=i+1}^\infty a_j) 
& = 2 \sum_{i=1}^N a_i (\sum_{j=i+1}^\infty a_j) + 2 \sum_{i=N+1}^\infty a_i (\sum_{j=i+1}^\infty a_j)\\
\end{align*}$$
Now it's obvious that the first term in this expression converges (it's a finite sum of convergent series) and the second term is bounded in the interval
$[-2\sum_{i=N+1}^\infty a_i, 2\sum_{i=N+1}^\infty a_i]$ but now we can simply let $N\to \infty$ and the afforementioned interval will approach the set $\{0\}$ and thus the second term also converges.
A: We have, obviously
$$
(a_1+a_2+\cdots+a_n)^2 = (a_1+a_2+\cdots+a_n)(a_1+a_2+\cdots+a_n)
$$
Now let's expand this
$$
\begin{align}
(a_1+\cdots+a_n)(a_1+\cdots+a_n)&=\\
&=a_1(a_1+\cdots+a_n)+\cdots+a_n(a_1+\cdots+a_n)\\
&= (a_1a_1+a_1a_2+\cdots) +\cdots+(a_na_1+\cdots+a_na_n)\\
&= \sum_{1\le i, j\le n} a_ia_j
\end{align}
$$
since we have all possible pairs $a_ja_j$. Let's break this sum into pieces 
First piece: Look at the pairs where $i=j$: you'll have
$$
\sum_{i=1}^n a_ia_i = \sum_{i=1}^n (a_i)^2
$$
Second piece: Now when $i\ne j$ we can group the terms in pairs, $a_ia_j=a_ja_i$. In other words, if we restrict ourselves to $1\le i < j\le n$, we'll have two terms for each choice of $i<j$ and thus we'll have the contribution of the remaining products,
$$
2 \sum_{1\le i < j\le n} a_ia_j
$$
which is exactly your second term. In other words
$$
(a_1+a_2+\cdots+a_n)^2 = \sum_{i=1}^n (a_i)^2 + 2\sum_{1\le i < j\le n} a_ia_j
$$ 

For your second question, as long as $n$ is finite, you'll of course always have a "last" term $2a_{n-1}a_n$, so there will be no collapse. Now suppose that $\langle a_i\rangle$ was an infinite sequence. Then we really mean that
$$
\begin{align}
\left(\sum_{i=1}^\infty a_i\right)^2 &= \lim_{n\rightarrow\infty}\left(\sum_{i=1}^n (a_i)^2 + 2\sum_{1\le i < j\le n} a_ia_j)\right)\\
&=\sum_{i=1}^\infty (a_i)^2 + 2\sum_{1\le i < j} a_ia_j
\end{align}
$$
and you see that there is no "last" term to collapse. Hope this helps.
