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If $a_k$ is a sequence of complex numbers, is it true that

$$\left|\sum_{k}a_k\right|\leq \sum_k |a_k|$$

Thank you

Chickenmancer
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user268193
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  • Yes. Basically, you can proceed inductively from $\lvert a+b\rvert\le\lvert a\rvert+\lvert b\rvert$ and then extend the inequality to the limit when the RHS is finite. The case when RHS diverges and LHS makes sense is trivially true. –  Aug 23 '16 at 11:08
  • Yes it is the generalisation of the triangle inequality. – StubbornAtom Aug 23 '16 at 11:10

2 Answers2

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Yes; this is called the triangle inequality, which is true for complex numbers (and can be visualized in the same way). If your sums are infinite, then the inequality still holds since the partial sums on the left are less than or equal to those on the right, so the limits also satisfy this inequality.

To prove the triangle inequality for $\Bbb{R}^n$, let $x,y\in \Bbb{R}^n$. Then $$||x+y||^2 = \langle x+y,x+y \rangle = \langle x,x\rangle+\langle y,y\rangle + 2\langle x,y \rangle \leq \langle x,x\rangle+\langle y,y\rangle + 2||x|| \cdot ||y|| = (||x||+||y||)^2$$

Where $||x||$ denote the norm ("absolute value") and $\langle x,y \rangle$ denotes the dot product of $x$ and $y$. The above inequality follows from the Cauchy-Schwarz-Bunyakovsky inequality, i.e that $\langle x,y \rangle \leq ||x||\cdot ||y||$.

florence
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First, read Prove the triangle inequality involving complex numbers.

Proof of the statement in question:

Let $(a_k)$ denote a sequence of complex numbers, and $s_n=\sum_{k=1}^n a_k$ be the $n$th partial sum.

One may observe, that for any $s_n$ we have

\begin{align*} |s_n| &= | a_1+\cdots +a_n|\\ & \leq |a_1 + \cdots +a_{n-1}| + |a_n|\\ &\vdots\\ &\leq |a_1|+\cdots+|a_n| \end{align*}

As a result of this being true, we have that $$\left|\sum_{k=1}^na_k \right| \leq \sum_{k=1}^n|a_k|$$ for all finite $n.$ By computing the limit,

$$\lim_{n\to\infty}\left|\sum_{k=1}^na_k \right| \leq \lim_{n\to\infty}\sum_{k=1}^n|a_k|$$ we get $$\left|\sum_{k=1}^\infty a_k \right| \leq \sum_{k=1}^\infty|a_k|.$$

And so the inequality holds for all finite sums, and infinite sums as well.

Chickenmancer
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