# Why is this proof attempt of "absolute convergence implies convergence" invalid?

I am self-teaching real analysis using this book and I try to not miss any proof. When I found the theorem of "absolute convergence implies convergence", I attempted to prove it as follows:

Let $$S_1$$ = $$\sum_{i=0}^n x_n$$ be a series, the convergence of which is our question. Then the corresponding sequence of its partial sums is {$${s_n}$$}.

By my understanding, we can investigate $$S_2$$ = $$\sum_{i=0}^n |x_n|$$ to assess whether or not $$S_1$$ is convergent. The sequence of partial sums for this series is then {$$|{s_n}|$$}.

A series is convergent if its corresponding sequence of partial sums is convergent. A sequence is convergent iff it is a Cauchy sequence, that is given $$e > 0$$ there is a number $$M$$ in real numbers such that if $$m,n$$ > $$M$$, $$|a_m - a_n| < e$$.

Plugging in terms for partial sums for $$S_2$$, we can say that: $$||s_m| - |s_n|| < e$$ (*), given that $$S_2$$ is convergent from the theorem statement.

Expanding the absolute value from (*), we can say that $$|s_m - s_n| < e$$ (which are equal to the partial sums of $$S_1$$) and conclude that $$S_1$$ is convergent.

Now I know that this is incorrect and in many ways not a proper proof. All proofs I have seen so far included either triangle inequality or a convergence test. In what statements is my induction wrong? I am trained in natural sciences and thus have trouble getting into mathematical formalism. Any help would be appreciated.

• It looks like you're arguing for the other direction, that convergence implies absolute convergence. (You start with $S_1$'s convergence and conclude $S_2$'s convergence.) Also, your notation is off: it should be e.g. $$\sum_{i=0}^\infty x_i$$ instead of $\sum_{i=0}^nx_n$. Apr 22 at 19:46
• Sorry, I'm not. I fill fix this asap. Thank you for your correction. Apr 22 at 19:49

You seem to have some misunderstandings about how absolute values work. The partial sums of $$S_2$$ are not necessarily equal to $$|s_n|$$. For instance, if $$x_0=1$$ and $$x_1=-1$$, then $$s_1=x_0+x_1=0$$ but the corresponding partial sum of $$S_2$$ is $$|x_0|+|x_1|=1+1=2$$. Similarly, it is not correct to "expand" $$||s_m|-|s_n||$$ to get $$|s_m-s_n|$$. If $$s_m$$ and $$s_n$$ have opposite signs, then $$||s_m|-|s_n||$$ will be different from $$|s_m-s_n|$$.