# Question on Cauchy Criterion of Series

We know that the Cauchy Criterion of a series is as follow (proof is taken as excerpt from an analysis book):

Theorem: A series $\sum_{j=1}^{\infty}a_j$ converges iff for all $\epsilon>0$ there is an $N\in \mathbb{N}$ so that for all $n\ge m \ge N$ we have $|\sum_{j=m}^{n} a_j|< \epsilon$.

Let $s_n$ denotes partial sum $\sum_{j=1}^n a_j$.

We know that the proof of $\Rightarrow$ direction makes use of the fact that convergent series implies the sequence of partial sum converges and thus is Cauchy in $\mathbb{R}$. So here is part of the proof, since sequence of partial sum is Cauchy, therefore given $\epsilon>0$, there exists $N \in \mathbb{N}$ such that for all $n \ge m \ge N$, $|s_n - s_{m-1}|< \epsilon$.

I do not get the $s_{m-1}$ part, i.e. I do not get why you can pick $m-1$, since there is possibility for $m=N$, and thus having $m-1<N$, which in this case you cannot guarantee the sequence of partial sum Cauchy.

1. Starting from the sequence of partial sum being Cauchy, $|s_n-s_m|< \epsilon$, for all $n,m\ge N$ (the definition of Cauchy sequence), and without loss of generality we can pick $n>m$ such that $|s_n- s_m|=|\sum_{j=m+1}^{n} a_j|< \epsilon$, but this differs from the statement of the theorem somehow.
2. I was also thinking of having $m \ne N$ and $m>N$ so that we have $m-1\ge N$ and that the original proof of the theorem that I showed right after the theorem is valid.
Pick your $N$ such that $n, m \geq N$ implies $|s_n - s_m| < \varepsilon$. Then use $N + 1$ instead. I agree that this is not symbol for symbol how one defines Cauchy, but playing little games with indices is something one has to get used to. – Dylan Moreland Jul 18 '12 at 21:52
It makes no difference at all if you use $n\geq m > N$ in the definition of Cauchy sequnces, so I guess your 2 option is a good solution to your problems. It is just a problem of what name you give to the numbers. If the sequence of partial sums is a Cauchy sequences there exists such $N$ as you defined. Now call $N' = N+1$ and conclude what you want.
Shouldn't it be there exists such $N'$ – Daniel Jul 18 '12 at 22:07