Baby Rudin theorem 3.45: Why is the first term of $|\sum_1^m{a_{n}}-\sum_1^n{a_{n}}|$? I have a rather stupid question regarding the terms of a series. So in Rudin's theorem 3.45, it says:

If $\sum{a_{n}}$ converges absolutely, them $\sum{a_{n}}$ converges.

And it proves as follow:

The assertion follows from the inequality
  $|\sum_{k=n}^{m}{a_{k}}|\le\sum_{k=n}^{m}|{a_{k}}|$  plus the cauchy
  criterion.

I know that by the Cauchy criterion, we have that for all $\xi>0$, there is a $N\inℕ$ such that $n,m>N$ implies that $|\sum_1^m{a_{n}}-\sum_1^n{a_{n}}|<\xi$. I have problem with the index here. I think $|\sum_1^m{a_{n}}-\sum_1^n{a_{n}}|=\sum_{n+1}^m{a_{n}}$, but Rudin's resultant partial sum is 
$\sum_{n}^m{a_{n}}$. Can anyone help me out here? I'd really appreciate it.
 A: 
I know that by the Cauchy criterion, we have that for all $\xi>0$, there is a $N\inℕ$ such that $n,m>N$ implies that $|\sum_1^m{a_{n}}-\sum_1^n{a_{n}}|<\xi$.

You've missed the absolute sign.  The premise is the convergence of $\sum_n |a_n|$, so Cauchy's criterion applied on the premise should gives
$$\forall \xi > 0: \exists N \in \Bbb{N}: \forall m,n > N: \left| \sum_{k=1}^m |a_k| - \sum_{k=1}^n |a_k| \right| <\xi.$$
Note that if $m > n$, the LHS of the inequality is a finite sum of $m - n$ terms.  The quoted book's inequality translated into words would read "the absolute value of a finite sum is less than or equal to a finite sum of absolute values".
Hence, for any $m,n > N$,
$$\left| \sum_{k=1}^m a_k - \sum_{k=1}^n a_k \right| = \left| \sum_{k=m+1}^n a_k \right| \le \left| \sum_{k=m+1}^n |a_k| \right| = \left| \sum_{k=1}^m |a_k| - \sum_{k=1}^n |a_k| \right| <\xi.$$

I think $|\sum_1^m{a_{n}}-\sum_1^n{a_{n}}|=\sum_{n+1}^m{a_{n}}$

Again, you're missing the absolute sign on the RHS.  Please take care of the indices as well.  The index for $(a_n)_n$ should be anything other than $m$ and $n$, which are used as the upper limit of the summations.
