> [!NOTE] Proposition ([[Absolutely Convergent Series|absolute convergence]] implies [[Convergent Real Series|convergence]]) > If $\sum_{n=1}^{\infty} |a_{n}| < \infty$, then $\sum_{n=1}^{\infty}a_{n} < \infty$. **Proof**: We define two new sequences with positive terms $\begin{gather} b_{n} =\begin{cases} a_{n} & a_{n} > 0, \\ 0 & a_{n} \leq 0; \end{cases} &c_{n}= \begin{cases} 0 & a_{n} \geq 0, \\ -a_{n} & a_{n} <0. \end{cases}\end{gather}$Note that $b_{n} \leq |a_{n}|$, $c_{n} \leq |a_{n}|,$ and $a_{n} = b_{n} -c_{n}.$ Since $\sum_{n=1}^{\infty} |a_{n}| < \infty$, It follows by [[Comparison Test for Series With Non-negative Terms (Corollary 1)]] that both $\sum_{n=1}^{\infty}b_{n}$ and $\sum_{n=1}^{\infty} c_{n}$ converge. Since $a_{n} =b_{n} - c_{n}$, it follows from [[Sum of Two Convergent Series is Convergent|sum rule for series]] that $\sum_{n=1}^{\infty} a_{n} < \infty$ > [!info] Note > Convergence ${\centernot\implies}$ Absolute Convergence. > For example, while $\sum_{n=1}^{\infty} \frac{1}{n} = \infty$ (The [[Alternating Harmonic Series|alternating harmonic series]] is convergent but not absolutely convergent i.e. [[Harmonic Numbers|the harmonic series]] is not convergent), $\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n} = \log 2$