Is it true that $\sum\limits_{i=1}^{n}\sum\limits_{i=1}^{n} a_i = \sum\limits_{i=1}^{n}a_i$, where $a_i\in \Bbb R$

Question

Is it true that $\sum\limits_{i=1}^{n}\sum\limits_{i=1}^{n} a_i = \sum\limits_{i=1}^{n}a_i$ where $a_i\in \Bbb R$ for $i=1,...n$.

I think it is true since $\sum\limits_{i=1}^{n} \left(\sum\limits_{i=1}^{n} a_i\right)=\sum\limits_{i=1}^{n} C =C$, where $C=\sum\limits_{i=1}^{n} a_i$.

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I wanted to use Cauchy's inequality in a proof and I was not sure if I could use this double summation. $$\sum\limits_{i=1}^{n}a_i^2\sum\limits_{i=1}^{n}b_i^2\sum\limits_{i=1}^{n}c_i^2 \geq \left(\sum\limits_{i=1}^{n}a_ib_i\right)^2\sum\limits_{i=1}^{n}c_k^2,$$ then I wanted to add a summation $\sum\limits_{i=1}^{n}$ to the right-hand side $$\sum\limits_{i=1}^{n}\left(\sum\limits_{i=1}^{n}a_ib_i\right)^2\sum\limits_{i=1}^{n}c_k^2\geq \sum\limits_{i=1}^{n}\sum\limits_{i=1}^{n}a_ib_ic_i = \sum\limits_{i=1}^{n}a_ib_ic_i.$$

Can I do that?

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Ok, I can't.

Answer 1

We have \begin{align*} \color{blue}{\sum_{i=1}^n\sum_{i=1}^na_i}&=\sum_{i=1}^n\left(\sum_{i=1}^na_i\right)\\ &=\sum_{i=1}^n\left(a_1+a_2+\cdots+a_n\right)\\ &=\left(a_1+a_2+\cdots+a_n\right)\sum_{i=1}^n1\\ &\,\,\color{blue}{=n\sum_{i=1}^na_i} \end{align*}

Note: A preferable notation for this situation is \begin{align*} \sum_{\color{blue}{j=1}}^{\color{blue}{n}}\sum_{i=1}^na_i=n\sum_{i=1}^na_i \end{align*}

Answer 2

No, it is not true since

\begin{equation*} \sum_{i=1}^n1=n, \end{equation*}

unless $\sum_{i=1}^na_i=0$.

Answer 3

It is not true, generally speaking. As Martin's comment points out: $$ \sum_{i=1}^n\sum_{i=1}^na_i=\sum_{i=1}^n\left(\sum_{i=1}^na_i\right)=\sum_{i=1}^nC=nC=n\sum_{i=1}^na_i $$ so, no. But, for instance, if $a_i=0\;\forall i$ then the equality you posted will be true. You can ask whether there are sequences $\{a\}_n$ such that your equality is true, or maybe prove that there is no possible $\{a\}_n$ (with at least one nonzero element) that satisfies your above equation.