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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.

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  • $\begingroup$ 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 $\sum_{i=0}^nx_n$. $\endgroup$ Apr 22 at 19:46
  • $\begingroup$ Sorry, I'm not. I fill fix this asap. Thank you for your correction. $\endgroup$
    – Ghostpunk
    Apr 22 at 19:49
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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 $|s_m-s_n|$.

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  • $\begingroup$ Thank you. This was one of the types of errors I was expecting. Thank you for your correction. $\endgroup$
    – Ghostpunk
    Apr 22 at 19:53

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