Prove if $\sum_{n=1}^\infty|a_n|<\infty$, then $\left|\sum_{n=1}^\infty a_n\right|\le \sum_{n=1}^\infty\left|a_n\right|$

Question

Prove if $\displaystyle\sum_{n=1}^\infty|a_n|<\infty$, then $\displaystyle\left|\sum_{n=1}^\infty a_n\right|\le \sum_{n=1}^\infty\left|a_n\right|$.

Upon initial observation, it seems like we should utilize the Triangle Inequality.

My initial thoughts:

PROOF: Since $\displaystyle \sum_{n=1}^\infty |a_n|$ is convergent and thus bounded, we then know the sequence of its partial sums is bounded (and that $\displaystyle \sum_{n=1}^\infty a_n$ converges since it converges absolutely, implying that its sequence of partial sums is also convergent and bounded). So, $$\left|\sum_{n = 1}^\infty a_n\right| = |a_1 + a_2 + a_3 + \cdots| \le |a_1| + |a_2| + |a_3| + \cdots = \sum_{n=1}^\infty|a_n|$$

Are we allowed to use the Triangle Inequality because the two series' sequences of partial sums are bounded? Or can we just use the Triangle Inequality straightaway?

Answer 1

No, it is not "allowed" to just use the triangle inequality like that. Indeed, you are using the very fact you want to prove, in your proof... The last inequality you wrote using "$\dots$" is exactly what you are trying to prove! This is called a circular proof and it's not really a proof at all.

Instead you need to be more careful. Let $S_N = \sum_{n=1}^N a_n$ be the sequence of partial sums. Using the actual triangle inequality and induction, you can prove that for all $N$, $$|S_N| \le \sum_{n=1}^N |a_n|.$$

Moreover, it is clear that $\sum_{n=1}^N |a_n| \le \sum_{n=1}^\infty |a_n|$, because the limit of a nondecreasing sequence is bigger than each individual term. Therefore you get that for all $N$, $$|S_N| \le \sum_{n=1}^\infty |a_n|.$$

Now if a sequence $u_n$ converges and for all $n$ you have $|u_n| \le C$, then $\lim_{n \to \infty} |u_n| \le C$. You can apply that to the sequence $\{S_N\}$ and $C = \sum_{n=1}^\infty |a_n|$ to finally get: $$\lim_{N \to \infty} |S_N| = \left| \sum_{n=1}^\infty a_n \right| \le \sum_{n=1}^\infty |a_n|.$$